problem solving involving law of sine

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Law of Sines Example Problem

The law of sines is a useful rule showing a relationship between an angle of a triangle and the length of the side opposite of the angle.

The law is expressed by the formula

The sine of the angle divided by the length of the opposite side is the same for every angle and its opposing side of the triangle.

Law of Sines – How does it work?

It is easy to show how this law works. First, let’s take the triangle from above and drop a vertical line to the side marked c .

This cuts the triangle into two right triangles which share a common side marked h.

The sine of an angle in a right triangle is the ratio of the length of the side opposite of the angle to the length of the hypotenuse of the right triangle. In other words:

Take the right triangle including the angle A . The length of the side opposite of A is h and the hypotenuse is equal to b .

Solve this for h and get

h = b sin A

Do the same thing for the right triangle including angle B . This time, the length of the side opposite of B is still h but the hypotenuse is equal to a .

h = a sin B

Since both of these equations are equal to h, they are equal to each other.

b sin A = a sin B

We can rewrite this to get the same letters on the same side of the equation to get

You can repeat process for every angle and get the same result. The overall result will be the law of sines.

Question: Use the law of sines to find the length of the side x.

Solution: The unknown side x is opposite the 46.5° angle and the side with length 7 is opposite the 39.4° angle. Plug these values into the Law of Sines equation.

Solve for x

7 sin(46.5°) = x sin(39.4°)

7 (0.725) = x (0.635)

5.078 = x (0.635)

Answer: The unknown side is equal to 8.

Bonus: If you wanted to find the missing angle and length of the last side of the triangle, remember that all three angles of a triangle all add up to 180°.

180° = 46.5° + 39.4° + C C = 94.1°

Use this angle in the law of sines the same way as above with either of the other angles and get a length of side c equal to 11.

Potential Issue of the Law of Sines

One potential problem to keep in mind using the law of sines is the possibility of two answers for an angle variable. This tends to appear when you are given two side values and an acute angle not between the two sides.

These two triangles are an example of this problem. The two sides are 100 and 75 in length and the 40° angle is not between these two sides. Notice how the side with length 75 could swing to hit a second place along the bottom side. Both of these angles will give a valid answer using the law of sines.

Fortunately, these two angle solutions add up to 180°. This is because the triangle formed by the two 75 sides is an isosceles triangle (triangle with two equal sides). The angles between the sides and their shared side will also be equal to each other. This means the angle on the other side of the angle θ will be the same as angle φ. The two angles added together make a straight line, or 180°.

Law of Sines Example Problem 2

Question : What are the two possible angles of a triangle with sides of 100 and 75 with a 40° as marked in the triangles above?

Solution : Use the law of sines formula where the 75 length is opposite of 40°, and 100 is opposite of θ.

sin θ = 0.857

θ + φ = 180°

φ = 180° – θ

φ = 180° – 58.97° φ = 121.03°

Answer : The two possible angles for this triangle is 58.97° and 121.03°.

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Great law of sines problems

These law of sines problems below will show you how to use the law of sines to solve some real life problems. You will need to use the sine formula shown below to solve these problems.

Law of sines

The ratio of the sine of an angle of a scalene triangle to the side opposite that angle is the same for all angles and sides in the triangle.

sin A / a = sin B / b =  sin C / c

Law of sines problems

Solving an angle-side-angle (asa) triangle with the law of sines.

Two fire-lookout stations are 15 miles apart, with station A directly east of station B. Both stations spot a fire. The angular direction of the fire from station B is N52°E and the angular direction of the fire from station A is N36°W. How far is the fire from station A? 

Solution The biggest trick is this problem is to understand the meaning of N52°E and N36°W.

N52°E means 52 degrees east of north.

N36°W means 36 degrees west of north.

Here is what the graph will look like after you draw it.

Notice that to find 38°, you need to subtract 52° from 90°. By the same token, to find 54°, you need to subtract 36° from 90°.

Notice also that the angle opposite 15 is missing, so we need to find it.

38 + 54 + missing angle = 180

92 + missing angle = 180, so missing angle  = 88

Now, we can use the sine rule  to find the distance the fire is from station A.

Let x be the distance from the fire to station A.

Solving an angle-angle-side (AAS) triangle with the law of sines

The leaning tower of pisa is inclined 5.5 degrees from the vertical. At a distance of 100 meters from the wall of the tower, the angle of elevation to the top is 30.5 degrees. Use the law of sines to estimate the height of the leaning tower (height as shown in the image below with the green line) 

The trickiest thing here is making the graph. We show it below. Notice that the height is shown with a green line.

Let us find angle y and angle z

angle y + 5.5 = 90, so angle y = 84.5

Notice that knowing the value of the angle y is not enough to find the unknown height of the leaning tower of pisa. You must know the value of the angle z!

84.5 + 30.5 + angle z =180

115 + angle z = 180

angle z =  65

Now, we can use law of sines.

Solving a side-side-angle (SSA) triangle with the ambiguous case of the law of sines

A homeowner is trying to build a raised garden bed that has a triangular shape. His neighbor gave him two pieces of lumber with lengths 20 feet and 8 feet and he puts them together to begin his triangle. He is planning to go to the store to buy the third piece of lumber to finish the triangle. He wants to build the triangular garden so that third piece of lumber will make an angle of 40 degrees with the piece that is 20 feet long. Will it be possible for him to make the garden? If so, how long should the third piece of lumber be?

If the homeowner could build such a garden, it will look like the figure shown above. An angle A would then be formed between the third piece of lumber and the piece that is 8 feet long.

Let us use the law of sines to see if this is possible.

sin (40 degrees) / 8 = sin (A) / 20

0.642 / 8 = sin (A) / 20

0.08025 = sin (A) / 20

Multiply both sides of the equation by 20

0.08025(20) = [sin (A) / 20](20)

0.08025(20) = sin (A) 

1.605 = sin(A)

The sine of an angle must be smaller than 1. Since Angle A cannot be formed, it will not be possible to make such a garden. Since the angle cannot be formed, the piece that is 8 feet long will never meet with the third piece. As a result, he will never know the length of the third piece. 

If the homeowner knows how to use the ambiguous case of law of sines , then he will do that first before driving to the store!

You cannot use the law of sines to solve word problems if the triangle you end up with is an SSS triangle or an SAS triangle. You must use the law of cosines instead,

Proof of the law of sines

Law of cosines

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Precalculus

Course: precalculus   >   unit 2.

• Trig word problem: stars
• General triangle word problems

Laws of sines and cosines review

Law of sines, law of cosines, practice set 1: solving triangles using the law of sines, example 1: finding a missing side, example 2: finding a missing angle.

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

Practice set 2: Solving triangles using the law of cosines

Example 1: finding an angle, example 2: finding a missing side, practice set 3: general triangle word problems, want to join the conversation.

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The Law of Sines

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10.1 Non-right Triangles: Law of Sines

Learning objectives.

In this section, you will:

• Use the Law of Sines to solve oblique triangles.
• Find the area of an oblique triangle using the sine function.
• Solve applied problems using the Law of Sines.

Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 1 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles .

Using the Law of Sines to Solve Oblique Triangles

In any triangle, we can draw an altitude , a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.

Any triangle that is not a right triangle is an oblique triangle . Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:

Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 5 .

Using the right triangle relationships, we know that sin α = h b sin α = h b and sin β = h a . sin β = h a . Solving both equations for h h gives two different expressions for h . h .

We then set the expressions equal to each other.

Similarly, we can compare the other ratios.

Collectively, these relationships are called the Law of Sines .

Note the standard way of labeling triangles: angle α α (alpha) is opposite side a ; a ; angle β β (beta) is opposite side b ; b ; and angle γ γ (gamma) is opposite side c . c . See Figure 6 .

While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.

Law of Sines

Given a triangle with angles and opposite sides labeled as in Figure 6 , the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways.

To solve an oblique triangle, use any pair of applicable ratios.

Solving for Two Unknown Sides and Angle of an AAS Triangle

Solve the triangle shown in Figure 7 to the nearest tenth.

The three angles must add up to 180 degrees. From this, we can determine that

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° α = 50° and its corresponding side a = 10. a = 10. We can use the following proportion from the Law of Sines to find the length of c . c .

Similarly, to solve for b , b , we set up another proportion.

Therefore, the complete set of angles and sides is

Solve the triangle shown in Figure 8 to the nearest tenth.

Using The Law of Sines to Solve SSA Triangles

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case . Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

Possible Outcomes for SSA Triangles

Oblique triangles in the category SSA may have four different outcomes. Figure 9 illustrates the solutions with the known sides a a and b b and known angle α . α .

Solving an Oblique SSA Triangle

Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth.

Use the Law of Sines to find angle β β and angle γ , γ , and then side c . c . Solving for β , β , we have the proportion

However, in the diagram, angle β β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β ? β ? Let’s investigate further. Dropping a perpendicular from γ γ and viewing the triangle from a right angle perspective, we have Figure 11 . It appears that there may be a second triangle that will fit the given criteria.

The angle supplementary to β β is approximately equal to 49.9°, which means that β = 180° − 49.9° = 130.1° . β = 180° − 49.9° = 130.1° . (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ , γ , we have

We can then use these measurements to solve the other triangle. Since γ ′ γ ′ is supplementary to the sum of α ′ α ′ and β ′ , β ′ , we have

Now we need to find c c and c ′ . c ′ .

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12 .

However, we were looking for the values for the triangle with an obtuse angle β . β . We can see them in the first triangle (a) in Figure 12 .

Given α = 80° , a = 120 , α = 80° , a = 120 , and b = 121 , b = 121 , find the missing side and angles. If there is more than one possible solution, show both.

Solving for the Unknown Sides and Angles of a SSA Triangle

In the triangle shown in Figure 13 , solve for the unknown side and angles. Round your answers to the nearest tenth.

In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle γ = 85° , γ = 85° , and its corresponding side c = 12 , c = 12 , and we know side b = 9. b = 9. We will use this proportion to solve for β . β .

To find β , β , apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for β . β . It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

In this case, if we subtract β β from 180°, we find that there may be a second possible solution. Thus, β = 180° − 48.3° ≈ 131.7° . β = 180° − 48.3° ≈ 131.7° . To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

which is impossible, and so β ≈ 48.3° . β ≈ 48.3° .

To find the remaining missing values, we calculate α = 180° − 85° − 48.3° ≈ 46.7° . α = 180° − 85° − 48.3° ≈ 46.7° . Now, only side a a is needed. Use the Law of Sines to solve for a a by one of the proportions.

The complete set of solutions for the given triangle is

Given α = 80° , a = 100 , b = 10 , α = 80° , a = 100 , b = 10 , find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

Finding the Triangles That Meet the Given Criteria

Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

Using the given information, we can solve for the angle opposite the side of length 10. See Figure 14 .

We can stop here without finding the value of α . α . Because the range of the sine function is [ − 1 , 1 ] , [ − 1 , 1 ] , it is impossible for the sine value to be 1.915. In fact, inputting sin − 1 ( 1.915 ) sin − 1 ( 1.915 ) in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.

Determine the number of triangles possible given a = 31 , a = 31 , b = 26 , b = 26 , β = 48° . β = 48° .

Finding the Area of an Oblique Triangle Using the Sine Function

Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as Area = 1 2 b h , Area = 1 2 b h , where b b is base and h h is height. For oblique triangles, we must find h h before we can use the area formula. Observing the two triangles in Figure 15 , one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property sin α = opposite hypotenuse sin α = opposite hypotenuse to write an equation for area in oblique triangles. In the acute triangle, we have sin α = h c sin α = h c or c sin α = h . c sin α = h . However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b b to form a right triangle. The angle used in calculation is α ′ , α ′ , or 180 − α . 180 − α .

Area of an Oblique Triangle

The formula for the area of an oblique triangle is given by

This is equivalent to one-half of the product of two sides and the sine of their included angle.

Finding the Area of an Oblique Triangle

Find the area of a triangle with sides a = 90 , b = 52 , a = 90 , b = 52 , and angle γ = 102° . γ = 102° . Round the area to the nearest integer.

Using the formula, we have

Find the area of the triangle given β = 42° , β = 42° , a = 7.2 ft , a = 7.2 ft , c = 3.4 ft . c = 3.4 ft . Round the area to the nearest tenth.

Solving Applied Problems Using the Law of Sines

The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.

Finding an Altitude

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16 . Round the altitude to the nearest tenth of a mile.

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a , a , and then use right triangle relationships to find the height of the aircraft, h . h .

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

The distance from one station to the aircraft is about 14.98 miles.

Now that we know a , a , we can use right triangle relationships to solve for h . h .

The aircraft is at an altitude of approximately 3.9 miles.

The diagram shown in Figure 17 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point B , B , is 62°, and the distance between the viewing points of the two end zones is 145 yards.

Access these online resources for additional instruction and practice with trigonometric applications.

• Law of Sines: The Basics
• Law of Sines: The Ambiguous Case

10.1 Section Exercises

Describe the altitude of a triangle.

Compare right triangles and oblique triangles.

When can you use the Law of Sines to find a missing angle?

In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator?

What type of triangle results in an ambiguous case?

For the following exercises, assume α α is opposite side a , β a , β is opposite side b , b , and γ γ is opposite side c . c . Solve each triangle, if possible. Round each answer to the nearest tenth.

α = 43° , γ = 69° , a = 20 α = 43° , γ = 69° , a = 20

α = 35° , γ = 73° , c = 20 α = 35° , γ = 73° , c = 20

α = 60° , α = 60° , β = 60° , β = 60° , γ = 60° γ = 60°

a = 4 , a = 4 , α = 60° , α = 60° , β = 100° β = 100°

b = 10 , b = 10 , β = 95° , γ = 30° β = 95° , γ = 30°

For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle A A is opposite side a , a , angle B B is opposite side b , b , and angle C C is opposite side c . c .

Find side b b when A = 37° , A = 37° , B = 49° , B = 49° , c = 5. c = 5.

Find side a a when A = 132° , C = 23° , b = 10. A = 132° , C = 23° , b = 10.

Find side c c when B = 37° , C = 21° , B = 37° , C = 21° , b = 23. b = 23.

For the following exercises, assume α α is opposite side a , β a , β is opposite side b , b , and γ γ is opposite side c . c . Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth.

α = 119° , a = 14 , b = 26 α = 119° , a = 14 , b = 26

γ = 113° , b = 10 , c = 32 γ = 113° , b = 10 , c = 32

b = 3.5 , b = 3.5 , c = 5.3 , c = 5.3 , γ = 80° γ = 80°

a = 12 , a = 12 , c = 17 , c = 17 , α = 35° α = 35°

a = 20.5 , a = 20.5 , b = 35.0 , b = 35.0 , β = 25° β = 25°

a = 7 , a = 7 , c = 9 , c = 9 , α = 43° α = 43°

a = 7 , b = 3 , β = 24° a = 7 , b = 3 , β = 24°

b = 13 , c = 5 , γ = 10° b = 13 , c = 5 , γ = 10°

a = 2.3 , c = 1.8 , γ = 28° a = 2.3 , c = 1.8 , γ = 28°

β = 119° , b = 8.2 , a = 11.3 β = 119° , b = 8.2 , a = 11.3

For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth.

Find angle A A when a = 24 , b = 5 , B = 22°. a = 24 , b = 5 , B = 22°.

Find angle A A when a = 13 , b = 6 , B = 20°. a = 13 , b = 6 , B = 20°.

Find angle B B when A = 12° , a = 2 , b = 9. A = 12° , a = 2 , b = 9.

For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth.

a = 5 , c = 6 , β = 35° a = 5 , c = 6 , β = 35°

b = 11 , c = 8 , α = 28° b = 11 , c = 8 , α = 28°

a = 32 , b = 24 , γ = 75° a = 32 , b = 24 , γ = 75°

a = 7.2 , b = 4.5 , γ = 43° a = 7.2 , b = 4.5 , γ = 43°

For the following exercises, find the length of side x . x . Round to the nearest tenth.

For the following exercises, find the measure of angle x , x , if possible. Round to the nearest tenth.

Notice that x x is an obtuse angle.

For the following exercise, solve the triangle. Round each answer to the nearest tenth.

For the following exercises, find the area of each triangle. Round each answer to the nearest tenth.

Find the radius of the circle in Figure 18 . Round to the nearest tenth.

Find the diameter of the circle in Figure 19 . Round to the nearest tenth.

Find m ∠ A D C m ∠ A D C in Figure 20 . Round to the nearest tenth.

Find A D A D in Figure 21 . Round to the nearest tenth.

Solve both triangles in Figure 22 . Round each answer to the nearest tenth.

Find A B A B in the parallelogram shown in Figure 23 .

Solve the triangle in Figure 24 . (Hint: Draw a perpendicular from H H to J K ). J K ). Round each answer to the nearest tenth.

Solve the triangle in Figure 25 . (Hint: Draw a perpendicular from N N to L M ). L M ). Round each answer to the nearest tenth.

In Figure 26 , A B C D A B C D is not a parallelogram. ∠ m ∠ m is obtuse. Solve both triangles. Round each answer to the nearest tenth.

Real-World Applications

A pole leans away from the sun at an angle of 7° 7° to the vertical, as shown in Figure 27 . When the elevation of the sun is 55° , 55° , the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth.

To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in Figure 28 . Determine the distance of the boat from station A A and the distance of the boat from shore. Round your answers to the nearest whole foot.

Figure 29 shows a satellite orbiting Earth. The satellite passes directly over two tracking stations A A and B , B , which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at A A and B B are measured to be 83.9° 83.9° and 86.2° , 86.2° , respectively. How far is the satellite from station A A and how high is the satellite above the ground? Round answers to the nearest whole mile.

A communications tower is located at the top of a steep hill, as shown in Figure 30 . The angle of inclination of the hill is 67° . 67° . A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is 16°. 16°. Find the length of the cable required for the guy wire to the nearest whole meter.

The roof of a house is at a 20° 20° angle. An 8-foot solar panel is to be mounted on the roof and should be angled 38° 38° relative to the horizontal for optimal results. (See Figure 31 ). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth.

Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be 37° 37° and 44° , 44° , as shown in Figure 32 . Find the distance of the plane from point A A to the nearest tenth of a kilometer.

A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in Figure 33 . Find the distance of the plane from point A A to the nearest tenth of a kilometer.

In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot.

In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot.

Points A A and B B are on opposite sides of a lake. Point C C is 97 meters from A . A . The measure of angle B A C B A C is determined to be 101°, and the measure of angle A C B A C B is determined to be 53°. What is the distance from A A to B , B , rounded to the nearest whole meter?

A man and a woman standing 3 1 2 3 1 2 miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot.

Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile.

A street light is mounted on a pole. A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot.

Three cities, A , B , A , B , and C , C , are located so that city A A is due east of city B . B . If city C C is located 35° west of north from city B B and is 100 miles from city A A and 70 miles from city B , B , how far is city A A from city B ? B ? Round the distance to the nearest tenth of a mile.

Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet.

Brian’s house is on a corner lot. Find the area of the front yard if the edges measure 40 and 56 feet, as shown in Figure 34 .

The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°.

A yield sign measures 30 inches on all three sides. What is the area of the sign?

Naomi bought a modern dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in Figure 35 .

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• Authors: Jay Abramson
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• Book title: Algebra and Trigonometry
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• Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/10-1-non-right-triangles-law-of-sines

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4.1.7: Trigonometry Word Problems

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Contextual use of triangle properties, ratios, theorems, and laws.

Angle of Depression and Angle of Elevation

One application of the trigonometric ratios is to find lengths that you cannot measure. Very frequently, angles of depression and elevation are used in these types of problems.

Angle of Depression: The angle measured down from the horizon or a horizontal line.

Angle of Elevation: The angle measured up from the horizon or a horizontal line.

What if you placed a ladder 10 feet from a haymow whose floor is 20 feet from the ground? How tall would the ladder need to be to reach the haymow's floor if it forms a \(30^{\circ}\) angle with the ground?

Example \(\PageIndex{1}\)

A math student is standing 25 feet from the base of the Washington Monument. The angle of elevation from her horizontal line of sight is \(87.4^{\circ}\). If her “eye height” is 5 ft, how tall is the monument?

We can find the height of the monument by using the tangent ratio.

\(\begin{aligned} \tan 87.4^{\circ} &=\dfrac{h}{25} \\ h&=25\cdot \tan 87.4^{\circ}=550.54 \end{aligned}\)

Adding 5 ft, the total height of the Washington Monument is 555.54 ft.

Example \(\PageIndex{2}\)

A 25 foot tall flagpole casts a 42 foot shadow. What is the angle that the sun hits the flagpole?

Draw a picture. The angle that the sun hits the flagpole is \(x^{\circ}\). We need to use the inverse tangent ratio.

\(\begin{aligned} \tan x &=\dfrac{42}{25} \\ \tan^{−1} \dfrac{42}{25}&\approx 59.2^{\circ}=x \end{aligned}\)

Example \(\PageIndex{3}\)

Elise is standing on top of a 50 foot building and sees her friend, Molly. If Molly is 30 feet away from the base of the building, what is the angle of depression from Elise to Molly? Elise’s eye height is 4.5 feet.

Because of parallel lines, the angle of depression is equal to the angle at Molly, or \(x^{\circ}\). We can use the inverse tangent ratio.

\(\tan^{−1} \left(\dfrac{54.5}{30}\right)=61.2^{\circ}=x\)

Example \(\PageIndex{4}\)

Mark is flying a kite and realizes that 300 feet of string are out. The angle of the string with the ground is \(42.5^{\circ}\). How high is Mark's kite above the ground?

It might help to draw a picture. Then write and solve a trig equation.

\(\begin{aligned} \sin 42.5^{\circ} &=\dfrac{x}{300}\\ 300\cdot \sin 42.5^{\circ} &=x \\ x&\approx 202.7\end{aligned}\)

The kite is about 202.7 feet off of the ground.

Example \(\PageIndex{5}\)

A 20 foot ladder rests against a wall. The base of the ladder is 7 feet from the wall. What angle does the ladder make with the ground?

It might help to draw a picture.

\(\begin{aligned} \cos x &=\dfrac{7}{20}\\ x&=\cos ^{−1}\dfrac{7}{20}\\ x&\approx 69.5^{\circ}\end{aligned}\)

Use what you know about right triangles to solve for the missing angle. If needed, draw a picture. Round all answers to the nearest tenth of a degree.

• A 75 foot building casts an 82 foot shadow. What is the angle that the sun hits the building?
• Over 2 miles (horizontal), a road rises 300 feet (vertical). What is the angle of elevation?
• A boat is sailing and spots a shipwreck 650 feet below the water. A diver jumps from the boat and swims 935 feet to reach the wreck. What is the angle of depression from the boat to the shipwreck?
• Standing 100 feet from the base of a building, Sam measures the angle to the top of the building from his eye height to be \(50^{\circ}\). If his eyes are 6 feet above the ground, how tall is the building?
• Over 4 miles (horizontal), a road rises 200 feet (vertical). What is the angle of elevation?
• A 90 foot building casts an 110 foot shadow. What is the angle that the sun hits the building?
• Luke is flying a kite and realizes that 400 feet of string are out. The angle of the string with the ground is \(50^{\circ}\). How high is Luke's kite above the ground?
• An 18 foot ladder rests against a wall. The base of the ladder is 10 feet from the wall. What angle does the ladder make with the ground?

Review (Answers)

To see the Review answers, open this PDF file and look for section 8.9.

Additional Resources

Interactive element.

Video: Trigonometry Word Problems Principles - Basic

Activities: Trigonometry Word Problems Discussion Questions

Practice: Trigonometry Word Problems

Real World: Measuring Mountains

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5.2: The Law of Sines

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• Carl Stitz & Jeff Zeager
• Lakeland Community College & Lorain County Community College

Trigonometry literally means ‘measuring triangles’ and with Chapter 10 under our belts, we are more than prepared to do just that. The main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is, find the length of each side of a triangle and the measure of each of its angles. In Sections 10.2 , 10.3 and 10.6 , we’ve had some experience solving right triangles. The following example reviews what we know.

Example 11.2.1

Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, find the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree.

For definitiveness, we label the triangle below.

To find the length of the missing side \(a\), we use the Pythagorean Theorem to get \(a^{2}+4^{2}=7^{2}\) which then yields \(a=\sqrt{33}\) units. Now that all three sides of the triangle are known, there are several ways we can find \(\alpha\) using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to \(\alpha\). According to Theorem 10.4 , \(\cos (\alpha)=\frac{4}{7}\). Since \(\alpha\) is an acute angle, \(\alpha=\arccos \left(\frac{4}{7}\right)\) radians. Converting to degrees, we find \(\alpha \approx 55.15^{\circ}\). Now that we have the measure of angle \(\alpha\), we could find the measure of angle \(\beta\) using the fact that \(\alpha\) and \(\beta\) are complements so \(\alpha+\beta=90^{\circ}\). Once again, we opt to use the data given to us in the problem. According to Theorem 10.4 , we have that \(\sin (\beta)=\frac{4}{7}\) so \(\beta=\arcsin \left(\frac{4}{7}\right)\) radians and we have \(\beta \approx 34.85^{\circ}\).

A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle 1 and the corresponding lowercase English letter represents the side 2 opposite that angle. Thus, \(a\) is the side opposite \(\alpha, b\) is the side opposite \(\beta\) and \(c\) is the side opposite \(\gamma\). Taken together, the pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it minimizes the chances of propagated error. 3 Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being. 4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases, we can use the Law of Sines to help.

Theorem 11.2. The Law of Sines

Given a triangle with angle-side opposite pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\), the following ratios hold

\(\frac{\sin (\alpha)}{a}=\frac{\sin (\beta)}{b}=\frac{\sin (\gamma)}{c}\)

or, equivalently,

\(\frac{a}{\sin (\alpha)}=\frac{b}{\sin (\beta)}=\frac{c}{\sin (\gamma)}\)

The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle \(\triangle A B C\) below, all of whose angles are acute, with angle-side opposite pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\). If we drop an altitude from vertex \(B\), we divide the triangle into two right triangles: \(\triangle A B Q\) and \(\triangle B C Q\). If we call the length of the altitude \(h\) (for height), we get from Theorem 10.4 that \(\sin (\alpha)=\frac{h}{c}\) and \(\sin (\gamma)=\frac{h}{a}\) so that \(h=c \sin (\alpha)=a \sin (\gamma)\). After some rearrangement of the last equation, we get \(\frac{\sin (\alpha)}{a}=\frac{\sin (\gamma)}{c}\). If we drop an altitude from vertex \(A\), we can proceed as above using the triangles \(\triangle A B Q\) and \(\triangle A C Q\) to get \(\frac{\sin (\beta)}{b}=\frac{\sin (\gamma)}{c}\), completing the proof for this case.

For our next case consider the triangle \(\triangle A B C\) below with obtuse angle \(\alpha\). Extending an altitude from vertex \(A\) gives two right triangles, as in the previous case: \(\triangle A B Q\) and \(\triangle A C Q\). Proceeding as before, we get \(h=b \sin (\gamma)\) and \(h=c \sin (\beta)\) so that \(\frac{\sin (\beta)}{b}=\frac{\sin (\gamma)}{c}\).

Dropping an altitude from vertex B also generates two right triangles, \(\triangle A B Q\) and \(\triangle B C Q\). We know that \(\sin \left(\alpha^{\prime}\right)=\frac{h^{\prime}}{c}\) so that \(h^{\prime}=c \sin \left(\alpha^{\prime}\right)\). Since \(\alpha^{\prime}=180^{\circ}-\alpha, \sin \left(\alpha^{\prime}\right)=\sin (\alpha)\), so in fact, we have \(h^{\prime}=c \sin (\alpha)\). Proceeding to \(\triangle B C Q\), we get \(\sin (\gamma)=\frac{h^{\prime}}{a} \text { so } h^{\prime}=a \sin (\gamma)\). Putting this together with the previous equation, we get \(\frac{\sin (\gamma)}{c}=\frac{\sin (\alpha)}{a}\), and we are finished with this case.

The remaining case is when \(\triangle A B C\) is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines.

Example 11.2.2

Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle.

• \(\alpha=120^{\circ}, a=7 \text { units, } \beta=45^{\circ}\)
• \(\alpha=85^{\circ}, \beta=30^{\circ}, c=5.25 \text { units }\)
• \(\alpha=30^{\circ}, a=1 \text { units, } c=4 \text { units }\)
• \(\alpha=30^{\circ}, a=2 \text { units, } c=4 \text { units }\)
• \(\alpha=30^{\circ}, a=3 \text { units, } c=4 \text { units }\)
• \(\alpha=30^{\circ}, a=4 \text { units, } c=4 \text { units }\)
• Knowing an angle-side opposite pair, namely \(\alpha\) and \(a\), we may proceed in using the Law of Sines. Since \(\beta=45^{\circ}\), we use \(\frac{b}{\sin \left(45^{\circ}\right)}=\frac{7}{\sin \left(120^{\circ}\right)}\) so \(b=\frac{7 \sin \left(45^{\circ}\right)}{\sin \left(120^{\circ}\right)}=\frac{7 \sqrt{6}}{3} \approx 5.72\) units. Now that we have two angle-side pairs, it is time to find the third. To find \(\gamma\), we use the fact that the sum of the measures of the angles in a triangle is \(180^{\circ}\). Hence, \(\gamma=180^{\circ}-120^{\circ}-45^{\circ}=15^{\circ}\). To find \(c\), we have no choice but to used the derived value \(\gamma=15^{\circ}\), yet we can minimize the propagation of error here by using the given angle-side opposite pair \((\alpha, a)\). The Law of Sines gives us \(\frac{c}{\sin \left(15^{\circ}\right)}=\frac{7}{\sin \left(120^{\circ}\right)}\) so that \(c=\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)} \approx 2.09\) units. 5

• Since we are given \((\alpha, a)\) and \(c\), we use the Law of Sines to find the measure of \(\gamma\). We start with \(\frac{\sin (\gamma)}{4}=\frac{\sin \left(30^{\circ}\right)}{1}\) and get \(\sin (\gamma)=4 \sin \left(30^{\circ}\right)=2\). Since the range of the sine function is [−1, 1], there is no real number with \(\sin (\gamma)=2\). Geometrically, we see that side \(a\) is just too short to make a triangle. The next three examples keep the same values for the measure of \(\alpha\) and the length of \(c\) while varying the length of \(a\). We will discuss this case in more detail after we see what happens in those examples.

Some remarks about Example 11.2.2 are in order. We first note that if we are given the measures of two of the angles in a triangle, say \(\alpha\) and \(\beta\), the measure of the third angle \(\gamma\) uniquely determined using the equation \(\gamma=180^{\circ}-\alpha-\beta\). Knowing the measures of all three angles of a triangle completely determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to find the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers 1 and 2 above. In number 1, the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case. 8 In number 2, the given side is adjacent to both angles which means we are in the so-called ‘Angle-Side-Angle’ (ASA) case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the ‘Angle-Side-Side’ (ASS) case. 9 In number 3, the length of the one given side \(a\) was too short to even form a triangle; in number 4, the length of a was just long enough to form a right triangle; in 5, \(a\) was long enough, but not too long, so that two triangles were possible; and in number 6, side \(a\) was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem.

Theorem 11.3

Suppose \((\alpha, a)\) and \((\gamma, c)\) are intended to be angle-side pairs in a triangle where \(\alpha\), \(a\) and \(c\) are given. Let \(h=c \sin (\alpha)\).

• If \(a<h\), then no triangle exists which satisfies the given criteria.
• If \(a=h\), then \(\gamma=90^{\circ}\) so exactly one (right) triangle exists which satisfies the criteria.
• If \(h<a<c\), then two distinct triangles exist which satisfy the given criteria.
• If \(a \geq c\), then \(\gamma\) is acute and exactly one triangle exists which satisfies the given criteria

Theorem 11.3 is proved on a case-by-case basis. If \(a<h\), then \(a<c \sin (\alpha)\). If a triangle were to exist, the Law of Sines would have \(\frac{\sin (\gamma)}{c}=\frac{\sin (\alpha)}{a}\) so that \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}>\frac{a}{a}=1\), which is impossible. In the figure below, we see geometrically why this is the case.

Simply put, if \(a<h\) the side \(a\) is too short to connect to form a triangle. This means if \(a \geq h\), we are always guaranteed to have at least one triangle, and the remaining parts of the theorem tell us what kind and how many triangles to expect in each case. If \(a = h\), then \(a=c \sin (\alpha)\) and the Law of Sines gives \(\frac{\sin (\alpha)}{a}=\frac{\sin (\gamma)}{c}\) so that \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}=\frac{a}{a}=1\). Here, \(\gamma=90^{\circ}\) as required. Moving along, now suppose \(h<a<c\). As before, the Law of Sines 10 gives \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}\). Since \(h<a\), \(c \sin (\alpha)<a\) or \(\frac{c \sin (\alpha)}{a}<1\) which means there are two solutions to \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}\): an acute angle which we’ll call \(\gamma_{0}\), and its supplement, \(180^{\circ}-\gamma_{0}\). We need to argue that each of these angles ‘fit’ into a triangle with \(\alpha\). Since \((\alpha, a)\) and \(\left(\gamma_{0}, c\right)\) are angle-side opposite pairs, the assumption \(c>a\) a in this case gives us \(\gamma_{0}>\alpha\). Since \(\gamma_{0}\) is acute, we must have that \(\alpha\) is acute as well. This means one triangle can contain both \(\alpha\) and \(\gamma_{0}\), giving us one of the triangles promised in the theorem. If we manipulate the inequality \(\gamma_{0}>\alpha\) a bit, we have \(180^{\circ}-\gamma_{0}<180^{\circ}-\alpha\) which gives \(\left(180^{\circ}-\gamma_{0}\right)+\alpha<180^{\circ}\). This proves a triangle can contain both of the angles \(\alpha\) and \(\left(180^{\circ}-\gamma_{0}\right)\), giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume \(a \geq c\). Then \(\alpha \geq \gamma\), which forces \(\gamma\) to be an acute angle. Hence, we get only one triangle in this case, completing the proof.

Example 11.2.3

Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is \(30^{\circ}\) and at the second point the angle is \(45^{\circ}\). Assuming a straight coastline, find the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point?

We sketch the problem below with the first observation point labeled as \(P\) and the second as \(Q\). In order to use the Law of Sines to find the distance \(d\) from \(Q\) to the island, we first need to find the measure of \(\beta\) which is the angle opposite the side of length 5 miles. To that end, we note that the angles \(\gamma\) and \(45^{\circ}\) are supplemental, so that \(\gamma=180^{\circ}-45^{\circ}=135^{\circ}\). We can now find \(\beta=180^{\circ}-30^{\circ}-\gamma=180^{\circ}-30^{\circ}-135^{\circ}=15^{\circ}\). By the Law of Sines, we have \(\frac{d}{\sin \left(30^{\circ}\right)}=\frac{5}{\sin \left(15^{\circ}\right)}\) which gives \(d=\frac{5 \sin \left(30^{\circ}\right)}{\sin \left(15^{\circ}\right)} \approx 9.66\) miles. Next, to find the point on the coast closest to the island, which we’ve labeled as \(C\), we need to find the perpendicular distance from the island to the coast. 11

Let \(x\) denote the distance from the second observation point \(Q\) to the point \(C\) and let \(y\) denote the distance from \(C\) to the island. Using Theorem 10.4 , we get \(\sin \left(45^{\circ}\right)=\frac{y}{d}\). After some rearranging, we find \(y=d \sin \left(45^{\circ}\right) \approx 9.66\left(\frac{\sqrt{2}}{2}\right) \approx 6.83\) miles. Hence, the island is approximately 6.83 miles from the coast. To find the distance from \(Q\) to \(C\), we note that \(\beta=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}\) so by symmetry, 12 we get \(x=y \approx 6.83\) miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point.

We close this section with a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise.

Theorem 11.4

Suppose \((\alpha, a),(\beta, b) \text { and }(\gamma, c)\) are the angle-side opposite pairs of a triangle. Then the area \(A\) enclosed by the triangle is given by

\(A=\frac{1}{2} b c \sin (\alpha)=\frac{1}{2} a c \sin (\beta)=\frac{1}{2} a b \sin (\gamma)\)

Example 11.2.4

Find the area of the triangle in Example 11.2.2 number 1.

From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose \(A=\frac{1}{2} a c \sin (\beta)\) from Theorem 11.4 because it uses the most pieces of given information. We are given \(a=7\) and \(\beta=45^{\circ}\), and we calculated \(c=\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\). Using these values, we find \(A=\frac{1}{2}(7)\left(\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\right) \sin \left(45^{\circ}\right)=\approx 5.18\) square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4 .

11.2.1 Exercises

In Exercises 1 - 20, solve for the remaining side(s) and angle(s) if possible. As in the text, \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\) are angle-side opposite pairs.

• \(\alpha=13^{\circ}, \beta=17^{\circ}, a=5\)
• \(\alpha=73.2^{\circ}, \beta=54.1^{\circ}, a=117\)
• \(\alpha=95^{\circ}, \beta=85^{\circ}, a=33.33\)
• \(\alpha=95^{\circ}, \beta=62^{\circ}, a=33.33\)
• \(\alpha=117^{\circ}, a=35, b=42\)
• \(\alpha=117^{\circ}, a=45, b=42\)
• \(\alpha=68.7^{\circ}, a=88, b=92\)
• \(\alpha=42^{\circ}, a=17, b=23.5\)
• \(\alpha=68.7^{\circ}, a=70, b=90\)
• \(\alpha=30^{\circ}, a=7, b=14\)
• \(\alpha=42^{\circ}, a=39, b=23.5\)
• \(\gamma=53^{\circ}, \alpha=53^{\circ}, c=28.01\)
• \(\alpha=6^{\circ}, a=57, b=100\)
• \(\gamma=74.6^{\circ}, c=3, a=3.05\)
• \(\beta=102^{\circ}, b=16.75, c=13\)
• \(\beta=102^{\circ}, b=16.75, c=18\)
• \(\beta=102^{\circ}, \gamma=35^{\circ}, b=16.75\)
• \(\beta=29.13^{\circ}, \gamma=83.95^{\circ}, b=314.15\)
• \(\gamma=120^{\circ}, \beta=61^{\circ}, c=4\)
• \(\alpha=50^{\circ}, a=25, b=12.5\)
• Find the area of the triangles given in Exercises 1, 12 and 20 above.

(Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10.6.6 ) in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 22 - 24, we first have you change road grades into angles and then use the Law of Sines in an application.

• Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a \(7 \%\) grade means that the road (hypotenuse) makes about a \(4^{\circ}\) angle with the horizontal. (It will not be exactly \(4^{\circ}\), but it’s pretty close.)
• What grade is given by a \(9.65^{\circ}\) angle made by the road and the horizontal? 13
• Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road. 14 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is \(6^{\circ}\). Use the Law of Sines to find the height of the tree. (Hint: First show that the tree makes a \(94^{\circ}\) angle with the road.)

(Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, \(\mathrm{N} 40^{\circ} \mathrm{E}\) (read "\(40^{\circ}\) east of north”) is a bearing which is rotated clockwise \(40^{\circ}\) from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of \(\theta=50^{\circ}\). Similarly, \(\mathrm{S} 50^{\circ} \mathrm{W}\) would point into Quadrant III along the terminal side of \(\theta=220^{\circ}\) because we started out pointing due south (along \(\theta=270^{\circ}\)) and rotated clockwise \(50^{\circ}\) back to \(220^{\circ}\). Counter-clockwise rotations would be found in the bearings \(\mathrm{N} 60^{\circ} \mathrm{W}\) (which is on the terminal side of \(\theta=150^{\circ}\)) and \(\mathrm{S} 27^{\circ} \mathrm{E}\) (which lies along the terminal side of \(\theta=297^{\circ}\)). These four bearings are drawn in the plane below.

The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.)

• \(\mathrm{S} 83^{\circ} \mathrm{E}\)
• \(\mathrm{N} 5.5^{\circ} \mathrm{E}\)
• \(\mathrm{N} 31.25^{\circ} \mathrm{W}\)
• \(\mathrm{S}{72}^{\circ} 41^{\prime} 12^{\prime \prime} \mathrm{W}\) 15
• \(\mathrm{N} 45^{\circ} \mathrm{E}\)
• \(\mathrm{S}{45}{ }^{\circ} \mathrm{W}\)
• The Colonel spots a campfire at a of bearing \(\mathrm{N} 42^{\circ} \mathrm{E}\) from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the fire to be \(\mathrm{N} 20^{\circ} \mathrm{W}\) from his current position. Determine the distance from the campfire to each man, rounded to the nearest foot.
• A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of \(\mathrm{N} 53^{\circ} \mathrm{W}\) which brings her to the Muffin Ridge Observatory. From there, she knows a bearing of \(\mathrm{S}6 5^{\circ} \mathrm{E}\) will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muffin Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead?
• The captain of the SS Bigfoot sees a signal flare at a bearing of \(\mathrm{N} 15^{\circ} \mathrm{E}\) from her current location. From his position, the captain of the HMS Sasquatch finds the signal flare to be at a bearing of \(\mathrm{N} 75^{\circ} \mathrm{W}\). If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is \(\mathrm{N} 50^{\circ} \mathrm{E}\), find the distances from the flare to each vessel, rounded to the nearest tenth of a mile.
• Carl spies a potential Sasquatch nest at a bearing of \(\mathrm{N} 10^{\circ} \mathrm{E}\) and radios Jeff, who is at a bearing of \(\mathrm{N} 50^{\circ} \mathrm{E}\) from Carl’s position. From Jeff’s position, the nest is at a bearing of \(\mathrm{S} 70^{\circ} \mathrm{W}\). If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot.
• A hiker determines the bearing to a lodge from her current position is \(\mathrm{S} 40^{\circ} \mathrm{W}\). She proceeds to hike 2 miles at a bearing of \(\mathrm{S} 20^{\circ} \mathrm{E}\) at which point she determines the bearing to the lodge is \(\mathrm{S}{75}{ }^{\circ} \mathrm{W}\). How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile.
• A watchtower spots a ship off shore at a bearing of \(\mathrm{N} 70^{\circ} \mathrm{E}\). A second tower, which is 50 miles from the first at a bearing of \(\mathrm{S} 80^{\circ} \mathrm{E}\) from the first tower, determines the bearing to the ship to be \(\mathrm{N} 25^{\circ} \mathrm{W}\). How far is the boat from the second tower? Round your answer to the nearest tenth of a mile.
• Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be \(75^{\circ}\) and radios Sally immediately to find the angle of inclination from her position to the craft is \(50^{\circ}\). How high off the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.)
• The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is \(55^{\circ}\). From a point five stories below the original observer, the angle of inclination to the gargoyle is \(20^{\circ}\). Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.)
• Prove that the Law of Sines holds when \(\triangle A B C\) is a right triangle.
• Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides.
• Discuss with your classmates why the Law of Sines cannot be used to find the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.)
• the information yields no triangle
• the information yields exactly one right triangle
• the information yields two distinct triangles
• the information yields exactly one obtuse triangle

Explain why you cannot choose \(a\) in such a way as to have \(\alpha=30^{\circ}, b=10\) and your choice of \(a\) yield only one triangle where that unique triangle has three acute angles.

• Use the cases and diagrams in the proof of the Law of Sines ( Theorem 11.2 ) to prove the area formulas given in Theorem 11.4 . Why do those formulas yield square units when four quantities are being multiplied together?

11.2.2 Answers

• \(\begin{array}{lll} \alpha=13^{\circ} & \beta=17^{\circ} & \gamma=150^{\circ} \\ a=5 & b \approx 6.50 & c \approx 11.11 \end{array}\)
• \(\begin{array}{lll} \alpha=73.2^{\circ} & \beta=54.1^{\circ} & \gamma=52.7^{\circ} \\ a=117 & b \approx 99.00 & c \approx 97.22 \end{array}\)
• Information does not produce a triangle
• \( \begin{array}{lll} \alpha=95^{\circ} & \beta=62^{\circ} & \gamma=23^{\circ} \\ a=33.33 & b \approx 29.54 & c \approx 13.07 \end{array}\)
• \(\begin{array}{lll} \alpha=117^{\circ} & \beta \approx 56.3^{\circ} & \gamma \approx 6.7^{\circ} \\ a=45 & b=42 & c \approx 5.89 \end{array}\)
• \(\begin{array}{lll} \alpha=68.7^{\circ} & \beta \approx 76.9^{\circ} & \gamma \approx 34.4^{\circ} \\ a=88 & b=92 & c \approx 53.36 \end{array}\)
• \(\begin{array}{lll} \alpha=42^{\circ} & \beta \approx 67.66^{\circ} & \gamma \approx 70.34^{\circ} \\ a=17 & b=23.5 & c \approx 23.93 \end{array}\)
• \(\begin{array}{lll} \alpha=30^{\circ} & \beta=90^{\circ} & \gamma=60^{\circ} \\ a=7 & b=14 & c=7 \sqrt{3} \end{array}\)
• \(\begin{array}{lll} \alpha=42^{\circ} & \beta \approx 23.78^{\circ} & \gamma \approx 114.22^{\circ} \\ a=39 & b=23.5 & c \approx 53.15 \end{array}\)
• \(\begin{array}{lll} \alpha=53^{\circ} & \beta=74^{\circ} & \gamma=53^{\circ} \\ a=28.01 & b \approx 33.71 & c=28.01 \end{array}\)
• \(\begin{array}{lll} \alpha=6^{\circ} & \beta \approx 169.43^{\circ} & \gamma \approx 4.57^{\circ} \\ a=57 & b=100 & c \approx 43.45 \end{array}\)
• \(\begin{array}{lll} \alpha \approx 78.59^{\circ} & \beta \approx 26.81^{\circ} & \gamma=74.6^{\circ} \\ a=3.05 & b \approx 1.40 & c=3 \end{array}\)
• \(\begin{array}{lll} \alpha \approx 28.61^{\circ} & \beta=102^{\circ} & \gamma \approx 49.39^{\circ} \\ a \approx 8.20 & b=16.75 & c=13 \end{array}\)
• \(\begin{array}{lll} \alpha=43^{\circ} & \beta=102^{\circ} & \gamma=35^{\circ} \\ a \approx 11.68 & b=16.75 & c \approx 9.82 \end{array}\)
• \(\begin{array}{lll} \alpha=66.92^{\circ} & \beta=29.13^{\circ} & \gamma=83.95^{\circ} \\ a \approx 593.69 & b=314.15 & c \approx 641.75 \end{array}\)
• \(\begin{array}{lll} \alpha=50^{\circ} & \beta \approx 22.52^{\circ} & \gamma \approx 107.48^{\circ} \\ a=25 & b=12.5 & c \approx 31.13 \end{array}\)

The area of the triangle from Exercise 12 is about 377.1 square units.

The area of the triangle from Exercise 20 is about 149 square units.

• \(\arctan \left(\frac{7}{100}\right) \approx 0.699 \text { radians }\), which is equivalent to \(\(4.004^{\circ}\)
• About 53 feet
• \(\theta=180^{\circ}\)
• \(\theta=353^{\circ}\)
• \(\theta=84.5^{\circ}\)
• \(\theta=270^{\circ}\)
• \(\theta=121.25^{\circ}\)
• \(\theta=197^{\circ} 18^{\prime} 48^{\prime \prime}\)
• \(\theta=45^{\circ}\)
• \(\theta=225^{\circ}\)

Sarge is about 2525 feet to the campfire.

The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles.

The HMS Sasquatch is about 2.9 miles from the flare.

• Jeff is about 371 feet from the nest.
• She is about 3.02 miles from the lodge
• The boat is about 25.1 miles from the second tower.
• The UFO is hovering about 9539 feet above the ground.

The gargoyle is about 27 feet from the observer on the lower floor.

The gargoyle is about 25 feet from the other building.

1 as well as the measure of said angle

2 as well as the length of said side

3 Your Science teachers should thank us for this.

4 Don’t worry! Radians will be back before you know it!

5 The exact value of \(\sin \left(15^{\circ}\right)\) could be found using the difference identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus “exact” here means \(\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\).

6 To find an exact expression for \(\beta\), we convert everything back to radians: \(\alpha=30^{\circ}=\frac{\pi}{6}\) radians, \(\gamma=\arcsin \left(\frac{2}{3}\right)\) radians and \(180^{\circ}=\pi\) radians. Hence, \(\beta=\pi-\frac{\pi}{6}-\arcsin \left(\frac{2}{3}\right)=\frac{5 \pi}{6}-\arcsin \left(\frac{2}{3}\right) \text { radians } \approx 108.19^{\circ}\).

7 An exact answer for \(\beta\) in this case is \(\beta=\arcsin \left(\frac{2}{3}\right)-\frac{\pi}{6} \text { radians } \approx 11.81^{\circ}\).

8 If this sounds familiar, it should. From high school Geometry, we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) and Side-Side-Side (SSS). If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that exactly one triangle exists which satisfies the given criteria.

9 In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case.

10 Remember, we have already argued that a triangle exists in this case!

11 Do you see why \(C\) must lie to the right of \(Q\)?

12 Or by Theorem 10.4 again . . .

13 I have friends who live in Pacifica, CA and their road is actually this steep. It’s not a nice road to drive.

14 The word ‘plumb’ here means that the tree is perpendicular to the horizontal.

15 See Example 10.1.1 in Section 10.1 for a review of the DMS system.

The Law of Sines

The Law of Sines (or Sine Rule ) is very useful for solving triangles:

a sin A = b sin B = c sin C

It works for any triangle:

And it says that:

When we divide side a by the sine of angle A it is equal to side b divided by the sine of angle B , and also equal to side c divided by the sine of angle C

Well, let's do the calculations for a triangle I prepared earlier:

a sin A = 8 sin(62.2°) = 8 0.885... = 9.04...

b sin B = 5 sin(33.5°) = 5 0.552... = 9.06...

c sin C = 9 sin(84.3°) = 9 0.995... = 9.04...

The answers are almost the same! (They would be exactly the same if we used perfect accuracy).

So now you can see that:

Is This Magic?

Not really, look at this general triangle and imagine it is two right-angled triangles sharing the side h :

The sine of an angle is the opposite divided by the hypotenuse, so:

a sin(B) and b sin(A) both equal h , so we get:

a sin(B) = b sin(A)

Which can be rearranged to:

a sin A = b sin B

We can follow similar steps to include c/sin(C)

How Do We Use It?

Let us see an example:

Example: Calculate side "c"

Now we use our algebra skills to rearrange and solve:

Finding an Unknown Angle

In the previous example we found an unknown side ...

... but we can also use the Law of Sines to find an unknown angle .

In this case it is best to turn the fractions upside down ( sin A/a instead of a/sin A , etc):

sin A a = sin B b = sin C c

Example: Calculate angle B

Sometimes there are two answers .

There is one very tricky thing we have to look out for:

Two possible answers.

This only happens in the " Two Sides and an Angle not between " case, and even then not always, but we have to watch out for it.

Just think "could I swing that side the other way to also make a correct answer?"

Example: Calculate angle R

The first thing to notice is that this triangle has different labels: PQR instead of ABC. But that's OK. We just use P,Q and R instead of A, B and C in The Law of Sines.

But wait! There's another angle that also has a sine equal to 0.9215...

The calculator won't tell you this but sin(112.9°) is also equal to 0.9215...

So, how do we discover the value 112.9°?

Easy ... take 67.1° away from 180°, like this:

180° − 67.1° = 112.9°

So there are two possible answers for R: 67.1° and 112.9° :

Both are possible! Each one has the 39° angle, and sides of 41 and 28.

So, always check to see whether the alternative answer makes sense.

• ... sometimes it will (like above) and there are two solutions
• ... sometimes it won't (like below) and there is one solution

For example this triangle from before.

As you can see, we can try swinging the "5.5" line around, but no other solution makes sense.

So this has only one solution.

Chapter 8: Applications of Trigonometry Functions

Section 8.2: the law of sines, learning outcomes.

By the end of this section, you will be able to:

• Use the Law of Sines to solve oblique triangles.
• Find the area of an oblique triangle using the sine function.
• Solve applied problems using the Law of Sines.

Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 1 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles .

Using the Law of Sines to Solve Obliques Triangles

In any triangle, we can draw an altitude , a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.

Any triangle that is not a right triangle is an oblique triangle . Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:

Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 5.

Using the right triangle relationships, we know that [latex]\sin \alpha =\frac{h}{b}[/latex] and [latex]\sin \beta =\frac{h}{a}[/latex]. Solving both equations for [latex]h[/latex] gives two different expressions for [latex]h[/latex].

We then set the expressions equal to each other.

Similarly, we can compare the other ratios.

Collectively, these relationships are called the Law of Sines .

Note the standard way of labeling triangles: angle [latex]\alpha [/latex] (alpha) is opposite side [latex]a[/latex]; angle [latex]\beta [/latex] (beta) is opposite side [latex]b[/latex]; and angle [latex]\gamma [/latex] (gamma) is opposite side [latex]c[/latex]. See Figure 6.

While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.

A General Note: Law of Sines

Given a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways.

[latex]\frac{\sin \alpha }{a}=\frac{\sin \beta }{b}=\frac{\sin \gamma }{c}[/latex]

[latex]\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }=\frac{c}{\sin \gamma }[/latex]

To solve an oblique triangle, use any pair of applicable ratios.

Example 1: Solving for Two Unknown Sides and Angle of an AAS Triangle

Solve the triangle shown in Figure 7 to the nearest tenth.

The three angles must add up to 180 degrees. From this, we can determine that

[latex]\begin{align} \beta =180^\circ -50^\circ -30^\circ =100^\circ \end{align}[/latex]

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle [latex]\alpha =50^\circ [/latex] and its corresponding side [latex]a=10[/latex]. We can use the following proportion from the Law of Sines to find the length of [latex]c[/latex].

[latex]\begin{align}&\frac{\sin \left(50^\circ \right)}{10}=\frac{\sin \left(30^\circ \right)}{c} \\ &c\frac{\sin \left(50^\circ \right)}{10}=\sin \left(30^\circ \right) && \text{Multiply both sides by }c. \\ &c=\sin \left(30^\circ \right)\frac{10}{\sin \left(50^\circ \right)} && \text{Multiply by the reciprocal to isolate }c. \\ &c\approx 6.5 \end{align}[/latex]

Similarly, to solve for [latex]b[/latex], we set up another proportion.

[latex]\begin{align} &\frac{\sin \left(50^\circ \right)}{10}=\frac{\sin \left(100^\circ \right)}{b} \\ &b\sin \left(50^\circ \right)=10\sin \left(100^\circ \right) && \text{Multiply both sides by }b. \\ &b=\frac{10\sin \left(100^\circ \right)}{\sin \left(50^\circ \right)} && \text{Multiply by the reciprocal to isolate }b. \\ &b\approx 12.9\end{align}[/latex]

Therefore, the complete set of angles and sides is

[latex]\begin{gathered} \alpha =50^\circ,\beta =100^\circ,\gamma =30^\circ \\ a=10,b\approx 12.9,c\approx 6.5 \end{gathered}[/latex]

Solve the triangle shown in Figure 8 to the nearest tenth.

[latex]\begin{gathered}\alpha ={98}^{\circ },\beta ={39}^{\circ },\gamma ={43}^{\circ } \\ a=34.6, b=22, c=23.8\end{gathered}[/latex]

Using The Law of Sines to Solve SSA Triangles

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case . Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

A General Note: Possible Outcomes for SSA Triangles

Oblique triangles in the category SSA may have four different outcomes. Figure 9 illustrates the solutions with the known sides [latex]a[/latex] and [latex]b[/latex] and known angle [latex]\alpha [/latex].

Example 2: Solving an Oblique SSA Triangle

Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth.

Use the Law of Sines to find angle [latex]\beta [/latex] and angle [latex]\gamma [/latex], and then side [latex]c[/latex]. Solving for [latex]\beta [/latex], we have the proportion

[latex]\begin{gathered} \frac{\sin \alpha }{a}=\frac{\sin \beta }{b} \\ \frac{\sin \left(35^\circ \right)}{6}=\frac{\sin \beta }{8}\\ \frac{8\sin \left(35^\circ \right)}{6}=\sin \beta \\ 0.7648\approx \sin \beta \\ {\sin }^{-1}\left(0.7648\right)\approx 49.9^\circ \\ \beta \approx 49.9^\circ \end{gathered}[/latex]

However, in the diagram, angle [latex]\beta [/latex] appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of [latex]\beta ?[/latex] Let’s investigate further. Dropping a perpendicular from [latex]\gamma [/latex] and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria.

The angle supplementary to [latex]\beta [/latex] is approximately equal to 49.9°, which means that [latex]\beta =180^\circ -49.9^\circ =130.1^\circ [/latex]. (Remember that the sine function is positive in both the first and second quadrants.) Solving for [latex]\gamma [/latex], we have

[latex]\gamma =180^\circ -35^\circ -130.1^\circ \approx 14.9^\circ [/latex]

We can then use these measurements to solve the other triangle. Since [latex]{\gamma }^{\prime }[/latex] is supplementary to [latex]\gamma [/latex], we have

[latex]{\gamma }^{\prime }=180^\circ -35^\circ -49.9^\circ \approx 95.1^\circ [/latex]

Now we need to find [latex]c[/latex] and [latex]{c}^{\prime }[/latex].

[latex]\begin{gathered}\frac{c}{\sin \left(14.9^\circ \right)}=\frac{6}{\sin \left(35^\circ \right)} \\ c=\frac{6\sin \left(14.9^\circ \right)}{\sin \left(35^\circ \right)}\approx 2.7 \end{gathered}[/latex]

[latex]\begin{gathered}\frac{{c}^{\prime }}{\sin \left(95.1^\circ \right)}=\frac{6}{\sin \left(35^\circ \right)} \\ {c}^{\prime }=\frac{6\sin \left(95.1^\circ \right)}{\sin \left(35^\circ \right)}\approx 10.4 \end{gathered}[/latex]

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12.

However, we were looking for the values for the triangle with an obtuse angle [latex]\beta [/latex]. We can see them in the first triangle (a) in Figure 12.

Given [latex]\alpha =80^\circ ,a=120[/latex], and [latex]b=121[/latex], find the missing side and angles. If there is more than one possible solution, show both.

Solution 1 [latex]\begin{align}&\alpha =80^\circ && a=120\hfill \\ &\beta \approx 83.2^\circ && b=121 \\ &\gamma \approx 16.8^\circ && c\approx 35.2 \end{align}[/latex] Solution 2 [latex]\begin{align}&{\alpha }^{\prime }=80^\circ &&{a}^{\prime }=120 \\ &{\beta }^{\prime }\approx 96.8^\circ &&{b}^{\prime }=121 \\ &{\gamma }^{\prime }\approx 3.2^\circ &&{c}^{\prime }\approx 6.8 \end{align}[/latex]

Example 3: Solving for the Unknown Sides and Angles of a SSA Triangle

In the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth.

In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle [latex]\gamma =85^\circ [/latex], and its corresponding side [latex]c=12[/latex], and we know side [latex]b=9[/latex]. We will use this proportion to solve for [latex]\beta [/latex].

[latex]\begin{align}\frac{\sin \left(85^\circ \right)}{12}&=\frac{\sin \beta }{9} && \text{Isolate the unknown}.\\ \frac{9\sin \left(85^\circ \right)}{12}&=\sin \beta \end{align}[/latex]

To find [latex]\beta [/latex], apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for [latex]\beta [/latex]. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

[latex]\begin{align}\beta &={\sin }^{-1}\left(\frac{9\sin \left(85^\circ \right)}{12}\right) \\ \beta &\approx {\sin }^{-1}\left(0.7471\right) \\ \beta &\approx 48.3^\circ \end{align}[/latex]

In this case, if we subtract [latex]\beta [/latex] from 180°, we find that there may be a second possible solution. Thus, [latex]\beta =180^\circ -48.3^\circ \approx 131.7^\circ [/latex]. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

[latex]\alpha =180^\circ -85^\circ -131.7^\circ \approx -36.7^\circ [/latex],

which is impossible, and so [latex]\beta \approx 48.3^\circ [/latex].

To find the remaining missing values, we calculate [latex]\alpha =180^\circ -85^\circ -48.3^\circ \approx 46.7^\circ [/latex]. Now, only side [latex]a[/latex] is needed. Use the Law of Sines to solve for [latex]a[/latex] by one of the proportions.

[latex]\begin{gathered} \frac{\sin \left(85^\circ \right)}{12}=\frac{\sin \left(46.7^\circ \right)}{a} \\ a\frac{\sin \left(85^\circ \right)}{12}=\sin \left(46.7^\circ \right) \\ a=\frac{12\sin \left(46.7^\circ \right)}{\sin \left(85^\circ \right)}\approx 8.8 \end{gathered}[/latex]

The complete set of solutions for the given triangle is

[latex]\begin{gathered} \alpha \approx 46.7^\circ \text{, }a\approx 8.8 \\ \beta \approx 48.3^\circ \text{, }b=9 \\ \gamma =85^\circ \text{, }c=12\end{gathered}[/latex]

Given [latex]\alpha =80^\circ ,a=100,b=10[/latex], find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

[latex]\beta \approx 5.7^\circ ,\gamma \approx 94.3^\circ ,c\approx 101.3[/latex]

Example 4: Finding the Triangles That Meet the Given Criteria

Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

Using the given information, we can solve for the angle opposite the side of length 10.

[latex]\begin{gathered}\frac{\sin \alpha }{10}=\frac{\sin \left(50^\circ \right)}{4} \\ \sin \alpha =\frac{10\sin \left(50^\circ \right)}{4} \\ \sin \alpha \approx 1.915 \end{gathered}[/latex]

We can stop here without finding the value of [latex]\alpha [/latex]. Because the range of the sine function is [latex]\left[-1,1\right][/latex], it is impossible for the sine value to be 1.915. In fact, inputting [latex]{\sin }^{-1}\left(1.915\right)[/latex] in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.

Determine the number of triangles possible given [latex]a=31,b=26,\beta =48^\circ [/latex].

Finding the Area of an Oblique Triangle Using the Sine Function

Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as [latex]\text{Area}=\frac{1}{2}bh[/latex], where [latex]b[/latex] is base and [latex]h[/latex] is height. For oblique triangles, we must find [latex]h[/latex] before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property [latex]\sin \alpha =\frac{\text{opposite}}{\text{hypotenuse}}[/latex] to write an equation for area in oblique triangles. In the acute triangle, we have [latex]\sin \alpha =\frac{h}{c}[/latex] or [latex]c\sin \alpha =h[/latex]. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base [latex]b[/latex] to form a right triangle. The angle used in calculation is [latex]{\alpha }^{\prime }[/latex], or [latex]180-\alpha [/latex].

A General Note: Area of an Oblique Triangle

The formula for the area of an oblique triangle is given by

[latex]\begin{align}\text{Area}&=\frac{1}{2}bc\sin \alpha \\ &=\frac{1}{2}ac\sin \beta \\ &=\frac{1}{2}ab\sin \gamma \end{align}[/latex]

This is equivalent to one-half of the product of two sides and the sine of their included angle.

Example 5: Finding the Area of an Oblique Triangle

Find the area of a triangle with sides [latex]a=90,b=52[/latex], and angle [latex]\gamma =102^\circ [/latex]. Round the area to the nearest integer.

Using the formula, we have

[latex]\begin{align}\text{Area}&=\frac{1}{2}ab\sin \gamma \\ \text{Area}&=\frac{1}{2}\left(90\right)\left(52\right)\sin \left(102^\circ \right) \\ \text{Area}&\approx 2289\text{square units} \end{align}[/latex]

Find the area of the triangle given [latex]\beta =42^\circ ,a=7.2\text{ft},c=3.4\text{ft}[/latex]. Round the area to the nearest tenth.

about 8.2 square feet

Solving Applied Problems Using the Law of Sines

The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.

Example 6: Finding an Altitude

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile.

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side [latex]a[/latex], and then use right triangle relationships to find the height of the aircraft, [latex]h[/latex].

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

[latex]\begin{gathered} \frac{\sin \left(130^\circ \right)}{20}=\frac{\sin \left(35^\circ \right)}{a} \\ a\sin \left(130^\circ \right)=20\sin \left(35^\circ \right) \\ a=\frac{20\sin \left(35^\circ \right)}{\sin \left(130^\circ \right)} \\ a\approx 14.98 \end{gathered}[/latex]

The distance from one station to the aircraft is about 14.98 miles.

Now that we know [latex]a[/latex], we can use right triangle relationships to solve for [latex]h[/latex].

[latex]\begin{gathered}\sin \left(15^\circ \right)=\frac{\text{opposite}}{\text{hypotenuse}} \\ \sin \left(15^\circ \right)=\frac{h}{a} \\ \sin \left(15^\circ \right)=\frac{h}{14.98} \\ h=14.98\sin \left(15^\circ \right) \\ h\approx 3.88 \end{gathered}[/latex]

The aircraft is at an altitude of approximately 3.9 miles.

The diagram shown in Figure 17 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point [latex]B[/latex], is 62°, and the distance between the viewing points of the two end zones is 145 yards.

Key Equations

Key concepts.

• The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
• According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
• There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution.
• The ambiguous case arises when an oblique triangle can have different outcomes.
• There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution.
• The Law of Sines can be used to solve triangles with given criteria.
• The general area formula for triangles translates to oblique triangles by first finding the appropriate height value.
• There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation.

Section 8.2 Homework Exercises

2. Compare right triangles and oblique triangles.

3. When can you use the Law of Sines to find a missing angle?

4. In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator?

5. What type of triangle results in an ambiguous case?

For the following exercises, assume [latex]\alpha [/latex] is opposite side [latex]a,\beta [/latex] is opposite side [latex]b[/latex], and [latex]\gamma [/latex] is opposite side [latex]c[/latex]. Solve each triangle, if possible. Round each answer to the nearest tenth.

6. [latex]\alpha =43^\circ ,\gamma =69^\circ ,a=20[/latex]

7. [latex]\alpha =35^\circ ,\gamma =73^\circ ,c=20[/latex]

8. [latex]\alpha =60^\circ ,\beta =60^\circ ,\gamma =60^\circ [/latex]

9. [latex]a=4,\alpha =60^\circ ,\beta =100^\circ [/latex]

10. [latex]b=10,\beta =95^\circ ,\gamma =30^\circ [/latex]

For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle [latex]A[/latex] is opposite side [latex]a[/latex], angle [latex]B[/latex] is opposite side [latex]b[/latex], and angle [latex]C[/latex] is opposite side [latex]c[/latex].

11. Find side [latex]b[/latex] when [latex]A=37^\circ ,B=49^\circ ,c=5[/latex].

12. Find side [latex]a[/latex] when [latex]A=132^\circ ,C=23^\circ ,b=10[/latex].

13. Find side [latex]c[/latex] when [latex]B=37^\circ ,C=21,b=23[/latex].

For the following exercises, assume [latex]\alpha [/latex] is opposite side [latex]a,\beta [/latex] is opposite side [latex]b[/latex], and [latex]\gamma [/latex] is opposite side [latex]c[/latex]. Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth.

14. [latex]\alpha =119^\circ ,a=14,b=26[/latex]

15. [latex]\gamma =113^\circ ,b=10,c=32[/latex]

16. [latex]b=3.5,c=5.3,\gamma =80^\circ [/latex]

17. [latex]a=12,c=17,\alpha =35^\circ [/latex]

18. [latex]a=20.5,b=35.0,\beta =25^\circ [/latex]

19. [latex]a=7,c=9,\alpha =43^\circ [/latex]

20. [latex]a=7,b=3,\beta =24^\circ [/latex]

21. [latex]b=13,c=5,\gamma =10^\circ [/latex]

22. [latex]a=2.3,c=1.8,\gamma =28^\circ [/latex]

23. [latex]\beta =119^\circ ,b=8.2,a=11.3[/latex]

For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth.

24. Find angle [latex]A[/latex] when [latex]a=24,b=5,B=22^\circ [/latex].

25. Find angle [latex]A[/latex] when [latex]a=13,b=6,B=20^\circ [/latex].

26. Find angle [latex]B[/latex] when [latex]A=12^\circ ,a=2,b=9[/latex].

For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth.

27. [latex]a=5,c=6,\beta =35^\circ [/latex]

28. [latex]b=11,c=8,\alpha =28^\circ [/latex]

29. [latex]a=32,b=24,\gamma =75^\circ [/latex]

30. [latex]a=7.2,b=4.5,\gamma =43^\circ [/latex]

For the following exercises, find the length of side [latex]x[/latex]. Round to the nearest tenth.

For the following exercises, find the measure of angle [latex]x[/latex], if possible. Round to the nearest tenth.

For the following exercises, find the area of each triangle. Round each answer to the nearest tenth.

66. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot.

67. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot.

68. Points [latex]A[/latex] and [latex]B[/latex] are on opposite sides of a lake. Point [latex]C[/latex] is 97 meters from [latex]A[/latex]. The measure of angle [latex]BAC[/latex] is determined to be 101°, and the measure of angle [latex]ACB[/latex] is determined to be 53°. What is the distance from [latex]A[/latex] to [latex]B[/latex], rounded to the nearest whole meter?

69. A man and a woman standing [latex]3\frac{1}{2}[/latex] miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot.

70. Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile.

71. A street light is mounted on a pole. A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot.

72. Three cities, [latex]A,B[/latex], and [latex]C[/latex], are located so that city [latex]A[/latex] is due east of city [latex]B[/latex]. If city [latex]C[/latex] is located 35° west of north from city [latex]B[/latex] and is 100 miles from city [latex]A[/latex] and 70 miles from city [latex]B[/latex], how far is city [latex]A[/latex] from city [latex]B?[/latex] Round the distance to the nearest tenth of a mile.

73. Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet.

75. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°.

76. A yield sign measures 30 inches on all three sides. What is the area of the sign?

• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution