problem solving with right triangles worksheet

Chapter 2: Trigonometric Ratios

Exercises: 2.3 Solving Right Triangles

Suggested Homework Problems

Exercises Homework 2.3

Exercise group.

For Problems 1–4, solve the triangle. Round answers to hundredths.

For Problems 5–10,

• Sketch the right triangle described.
• Solve the triangle.

[latex]A = 42{^o}, c = 26[/latex]

[latex]B = 28{^o}, c = 6.8[/latex]

[latex]B = 33{^o}, a = 300[/latex]

[latex]B = 79{^o}, a = 116[/latex]

[latex]A = 12{^o}, a = 4[/latex]

[latex]A = 50{^o}, a = 10[/latex]

For Problems 11–16,

• Without doing the calculations, list the steps you would use to solve the triangle.

[latex]B = 53.7{^o}, b = 8.2[/latex]

[latex]B = 80{^o}, a = 250[/latex]

[latex]A = 25{^o}, b = 40[/latex]

[latex]A = 15{^o}, c = 62[/latex]

[latex]A = 64.5{^o}, c = 24[/latex]

[latex]B = 44{^o}, b = 0.6[/latex]

For Problems 17–22, find the labeled angle. Round your answer to tenths of a degree.

For Problems 23–28, evaluate the expression and sketch a right triangle to illustrate.

[latex]\sin^{-1} 0.2[/latex]

[latex]\cos^{-1} 0.8[/latex]

[latex]\tan^{-1} 1.5[/latex]

[latex]\tan^{-1} 2.5[/latex]

[latex]\cos^{-1} 0.2839[/latex]

[latex]\sin^{-1} 0.4127[/latex]

For Problems 29–32, write two different equations for the statement.

The cosine of [latex]15 {^o}[/latex] is [latex]0.9659\text{.}[/latex]

The sine of [latex]70 {^o}[/latex] is [latex]0.9397\text{.}[/latex]

The angle whose tangent is [latex]3.1445[/latex] is [latex]65 {^o}\text{.}[/latex]

The angle whose cosine is [latex]0.0872[/latex] is [latex]85 {^o}\text{.}[/latex]

Evaluate the expressions and explain what each means. [latex]\begin{equation*} \sin^{-1} (0.6), (\sin 6{^o})^{-1} \end{equation*}[/latex]

Evaluate the expressions and explain what each means. [latex]\begin{equation*} \cos^{-1} (0.36), (\cos 36{^o})^{-1} \end{equation*}[/latex]

For Problems 35–38,

• Sketch a right triangle that illustrates the situation. Label your sketch with the given information.
• Choose the appropriate trig ratio and write an equation, then solve the problem.

The gondola cable for the ski lift at Snowy Peak is [latex]2458[/latex] yards long and climbs [latex]1860[/latex] feet. What angle with the horizontal does the cable make?

The Leaning Tower of Pisa is [latex]55[/latex] meters in length. An object dropped from the top of the tower lands [latex]4.8[/latex] meters from the base of the tower. At what angle from the horizontal does the tower lean?

A mining company locates a vein of minerals at a depth of [latex]32[/latex] meters. However, there is a layer of granite directly above the minerals, so they decide to drill at an angle, starting [latex]10[/latex] meters from their original location. At what angle from the horizontal should they drill?

The birdhouse in Carolyn’s front yard is [latex]12[/latex] feet tall, and its shadow at [latex]4[/latex] pm is [latex]15[/latex] feet [latex]4[/latex] inches long. What is the angle of elevation of the sun at [latex]4[/latex] pm?

For Problems 39–42,

[latex]a = 18, b = 26[/latex]

[latex]a = 35, b = 27[/latex]

[latex]b = 10.6 , c = 19.2[/latex]

[latex]a = 88, c = 132[/latex]

For Problems 43–48,

• Make a sketch that illustrates the situation. Label your sketch with the given information.
• Write an equation and solve the problem.

The Mayan pyramid of El Castillo at Chichén Itzá in Mexico has [latex]91[/latex] steps. Each step is 26 cm high and 30 cm deep.

• What angle does the side of the pyramid make with the horizontal?
• What is the distance up the face of the pyramid, from base to top platform?

An airplane begins its descent when its altitude is 10 kilometers. The angle of descent should be [latex]3{^o}[/latex] from horizontal.

• How far from the airport (measured along the ground) should the airplane begin its descent?
• How far will the airplane travel on its descent to the airport?

A communications satellite is in a low earth orbit (LOE) at an altitude of 400 km. From the satellite, the angle of depression to earth’s horizon is [latex]19.728{^o}\text{.}[/latex] Use this information to calculate the radius of the earth.

The first Ferris wheel was built for the [latex]1893[/latex] Chicago World’s Fair. It had a diameter of [latex]250[/latex] feet, and the boarding platform, at the base of the wheel, was [latex]14[/latex] feet above the ground. If you boarded the wheel and rotated through an angle of [latex]50{^o}\text{,}[/latex] what would be your height above the ground?

To find the distance across a ravine, Delbert takes some measurements from a small airplane. When he is a short distance from the ravine at an altitude of [latex]500[/latex] feet, he finds that the angle of depression to the near side of the ravine is [latex]56{^o}\text{,}[/latex] and the angle of depression to the far side is [latex]32{^o}\text{.}[/latex] What is the width of the ravine? (Hint: First find the horizontal distance from Delbert to the near side of the ravine.)

The window in Francine’s office is [latex]4[/latex] feet wide and [latex]5[/latex] feet tall. The bottom of the window is 3 feet from the floor. When the sun is at an angle of elevation of [latex]64{^o}\text{,}[/latex] what is the area of the sunny spot on the floor?

Which of the following numbers are equal to [latex]\cos 45{^o}\text{?}[/latex]

• [latex]\displaystyle \dfrac{\sqrt{2}}{2}[/latex]
• [latex]\displaystyle \dfrac{1}{\sqrt{2}}[/latex]
• [latex]\displaystyle \dfrac{2}{\sqrt{2}}[/latex]
• [latex]\displaystyle \sqrt{2}[/latex]

Which of the following numbers are equal to [latex]\tan 30{^o}\text{?}[/latex]

• [latex]\displaystyle \sqrt{3}[/latex]
• [latex]\displaystyle \dfrac{1}{\sqrt{3}}[/latex]
• [latex]\displaystyle \dfrac{\sqrt{3}}{3}[/latex]
• [latex]\displaystyle \dfrac{3}{\sqrt{3}}[/latex]

Which of the following numbers are equal to [latex]\tan 60{^o}\text{?}[/latex]

Which of the following numbers are equal to [latex]\sin 60{^o}\text{?}[/latex]

• [latex]\displaystyle \dfrac{3}{\sqrt{2}}[/latex]
• [latex]\displaystyle \dfrac{\sqrt{3}}{2}[/latex]
• [latex]\displaystyle \dfrac{\sqrt{2}}{3}[/latex]
• [latex]\displaystyle \dfrac{2}{\sqrt{3}}[/latex]

For Problems 53–58, choose all values from the list below that are exactly equal to, or decimal approximations for, the given trig ratio. (Try not to use a calculator!)

[latex]\cos 30{^o}[/latex]

[latex]\sin 45{^o}[/latex]

[latex]\tan 30{^o}[/latex]

[latex]\cos 60{^o}[/latex]

[latex]\sin 90{^o}[/latex]

[latex]\cos 0{^o}[/latex]

Fill in the table from memory with exact values. Do you notice any patterns that might help you memorize the values?

Fill in the table from memory with decimal approximations to four places.

For Problems 61 and 62, compare the given value with the trig ratios of the special angles to answer the questions. Try not to use a calculator.

Is the acute angle larger or smaller than [latex]45{^o}\text{?}[/latex]

• [latex]\displaystyle \sin \alpha = 0.7[/latex]
• [latex]\displaystyle \tan \beta = 1.2[/latex]
• [latex]\displaystyle \cos \gamma = 0.65[/latex]

Is the acute angle larger or smaller than [latex]60{^o}\text{?}[/latex]

• [latex]\displaystyle \cos \theta = 0.75[/latex]
• [latex]\displaystyle \tan \phi = 1.5[/latex]
• [latex]\displaystyle \sin \psi = 0.72[/latex]

For Problems 63–72, solve the triangle. Give your answers as exact values.

• Find the perimeter of a regular hexagon if the apothegm is [latex]8[/latex] cm long. (The apothegm is the segment from the center of the hexagon and perpendicular to one of its sides.)
• Find the area of the hexagon.

Triangle [latex]ABC[/latex] is equilateral, and its angle bisectors meet at point [latex]P\text{.}[/latex] The sides of [latex]\triangle ABC[/latex] are 6 inches long. Find the length of [latex]AP\text{.}[/latex]

Find an exact value for the area of each triangle.

Find an exact value for the perimeter of each parallelogram.

• Find the area of the outer square.
• Find the dimensions and the area of the inner square.
• What is the ratio of the area of the outer square to the area of the inner square?
• Find the area of the inner square.
• Find the dimensions and the area of the outer square.

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Right-Triangle Word Problems

What is a right-triangle word problem.

A right-triangle word problem is one in which you are given a situation (like measuring something's height) that can be modelled by a right triangle. You will draw the triangle, label it, and then solve it; finally, you interpret this solution within the context of the original exercise.

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Right Triangle Word Problems

Once you've learned about trigonometric ratios (and their inverses), you can solve triangles. Naturally, many of these triangles will be presented in the context of word problems. A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it. Once you've got a helpful diagram, the math is usually pretty straightforward.

• A six-meter-long ladder leans against a building. If the ladder makes an angle of 60° with the ground, how far up the wall does the ladder reach? How far from the wall is the base of the ladder? Round your answers to two decimal places, as needed.

First, I'll draw a picture. It doesn't have to be good or to scale; it just needs to be clear enough that I can keep track of what I'm doing. My picture is:

To figure out how high up the wall the top of the ladder is, I need to find the height h of my triangle.

Since they've given me an angle measure and "opposite" and the hypotenuse for this angle, I'll use the sine ratio for finding the height:

sin(60°) = h/6

6 sin(60°) = h = 3sqrt[3]

Plugging this into my calculator, I get an approximate value of 5.196152423 , which I'll need to remember to round when I give my final answer.

For the base, I'll use the cosine ratio:

cos(60°) = b/6

6×cos(60°) = b = 3

Nice! The answer is a whole number; no radicals involved. I won't need to round this value when I give my final answer. Checking the original exercise, I see that the units are "meters", so I'll include this unit on my numerical answers:

ladder top height: about 5.20 m

ladder base distance: 3 m

Note: Unless you are told to give your answer in decimal form, or to round, or in some other way not to give an "exact" answer, you should probably assume that the "exact" form is what they're wanting. For instance, if they hadn't told me to round my numbers in the exercise above, my value for the height would have been the value with the radical.

• A five-meter-long ladder leans against a wall, with the top of the ladder being four meters above the ground. What is the approximate angle that the ladder makes with the ground? Round to the nearest whole degree.

As usual, I'll start with a picture, using "alpha" to stand for the base angle:

They've given me the "opposite" and the hypotenuse, and asked me for the angle value. For this, I'll need to use inverse trig ratios.

sin(α) = 4/5

m(α) = sin −1 (4/5) = 53.13010235...

(Remember that m(α) means "the measure of the angle α".)

So I've got a value for the measure of the base angle. Checking the original exercise, I see that I am supposed to round to the nearest whole degree, so my answer is:

base angle: 53°

• You use a transit to measure the angle of the sun in the sky; the sun fills 34' of arc. Assuming the sun is 92,919,800 miles away, find the diameter of the sun. Round your answer to the nearest whole mile.

First, I'll draw a picture, labelling the angle on the Earth as being 34 minutes, where minutes are one-sixtieth of a degree. My drawing is *not* to scale!:

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Hmm... This "ice-cream cone" picture doesn't give me much to work with, and there's no right triangle.

The two lines along the side of my triangle measure the lines of sight from Earth to the sides of the Sun. What if I add another line, being the direct line from Earth to the center of the Sun?

Now that I've got this added line, I have a right triangle — two right triangles, actually — but I only need one. I'll use the triangle on the right.

(The angle measure , "thirty-four arc minutes", is equal to 34/60 degrees. Dividing this in half is how I got 17/60 of a degree for the smaller angle.)

I need to find the width of the Sun. That width will be twice the base of one of the right triangles. With respect to my angle, they've given me the "adjacent" and have asked for the "opposite", so I'll use the tangent ratio:

tan(17/60°) = b/92919800

92919800×tan(17/60°) = b = 459501.4065...

This is just half the width; carrying the calculations in my calculator (to minimize round-off error), I get a value of 919002.8129 . This is higher than the actual diameter, which is closer to 864,900 miles, but this value will suffice for the purposes of this exercise.

diameter: about 919,003 miles

• A private plane flies 1.3 hours at 110 mph on a bearing of 40°. Then it turns and continues another 1.5 hours at the same speed, but on a bearing of 130°. At the end of this time, how far is the plane from its starting point? What is its bearing from that starting point? Round your answers to whole numbers.

The bearings tell me the angles from "due north", in a clockwise direction. Since 130 − 40 = 90 , these two bearings create a right angle where the plane turns. From the times and rates, I can find the distances travelled in each part of the trip:

1.3 × 110 = 143 1.5 × 110 = 165

Now that I have the lengths of the two legs, I can set up a triangle:

(The angle θ is the bearing, from the starting point, of the plane's location at the ending point of the exercise.)

I can find the distance between the starting and ending points by using the Pythagorean Theorem :

143 2 + 165 2 = c 2 20449 + 27225 = c 2 47674 = c 2 c = 218.3437657...

The 165 is opposite the unknown angle, and the 143 is adjacent, so I'll use the inverse of the tangent ratio to find the angle's measure:

165/143 = tan(θ)

tan −1 (165/143) = θ = 49.08561678...

But this angle measure is not the "bearing" for which they've asked me, because the bearing is the angle with respect to due north. To get the measure they're wanting, I need to add back in the original forty-degree angle:

distance: 218 miles

bearing: 89°

Related: Another major class of right-triangle word problems you will likely encounter is angles of elevation and declination .

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Working with Right Triangles Worksheets

This is the one of the geometric shapes that literally makes the construction of perfectly straight edges possible. Using the Pythagorean theorem, we can find missing sides of these shapes. As we advance into trigonometry, we will learn that you do not much information at all about these shapes to be able to determine all of the angles and lengths of sides. Something that people often do not realize is that if put a diagonal line across a rectangle and cut along that line, you have just created two right-angled triangles. Contractors use the property of geometry all day to create perfectly level and straight structures. This series of lessons and worksheets teach students how to complete figure out all measures on right triangles through a wide variety of techniques.

Aligned Standard: HSG-SRT.C.8

• Find Opposite Step-by-Step Lesson - Given two sides, find the third of the right triangle.
• Guided Lesson - In summer, I always think of problems in my head like number two.
• Guided Lesson Explanation - These problems real help students get the concepts that are mystifying to lower level geometry students.
• Practice Worksheet - This one took me a while. All nicely thought out real world problem set for you to work with.
• Matching Worksheet - In real life if you took the time to think and solve problems like this, you would need to find a hobby.
• Working with Right Triangles Worksheet Five Pack - All these problems have to visualized because they are in sentences.
• Answer Keys - These are for all the unlocked materials above.

Homework Sheets

There are just so many different ways you can handle these problems.

• Homework 1 - A triangle contains exactly one 90° angle. The other two angles must total exactly 90 degrees. The famous Pythagoras Theorem defines the relationship between the three sides of a right triangle.
• Homework 2 - Jack saw a building that is 75 feet in height. The building casts a 30 foot shadow. What is the angle of elevation from the end of the shadow to the top of the building with respect to the ground?
• Homework 3 - A right triangle's opposite side is 10 and the hypotenuse is 18. Find the value of normal base?

Practice Worksheets

Over the course of the practice I show two different methods for answering the problems.

• Practice 1 - Find the value of normal opposite side?
• Practice 2 - Mark drives 10 km due east of his home. Then he heads 12km north. What is the total distance that he has travelled from his house?
• Practice 3 - Mr. Mike wants to purchase a square shaped table for his office. Table height is 6 feet. Mr. Mike ties a diagonal ribbon on the table. How long must be the ribbon?

Math Skill Quizzes

The first one is just with plain old triangles. The last two are all application word problems.

• Quiz 1 - A right triangle has a opposite side that is 10 units in length and a hypotenuse that is 14 units. Find the value of base?
• Quiz 2 - A road light is 40 feet in height and casts a shadow that is 15 feet long. What is the angle of elevation from the end of the shadow to the top of the road light with respect to the ground? This is assuming that the road light is 100% straight.
• Quiz 3 - Alan has a square shaped television. The height of television is 8 inches. Alan wants to put a diagonal paper on the television. How long must the paper be?

Tips for Working with Right Triangles

A right-angled triangle is one where one of the angles is 90 degrees. As you can see in the diagram to the right, that angle is often denoted by the presence of a square. There are a few unique features of these geometric shapes that you should be aware of. The longest side that is formed as a result of the right angle is opposite that right angle and called the hypotenuse. The altitude in a two-dimensional right triangle is the side that indicates height. The bottom side is referred to as the base. Depending on the orientation the two sides that are not the hypotenuse are interchangeable as result these sides are often referred to as legs. Since a triangle contains 180 degrees of internal angles, the sum of the measures of the legs must be 90 degrees.

The angles are often labeled which allows you to name the triangles. When naming triangles, we can pick any angle name them by labelling the angles in either a clockwise or counterclockwise direction. It does not which you pick as long as the angles fall in same consecutive order that they are found.

Working with right-angled triangles is relatively easier compared to other types of triangles. The reason behind this is simple. You always know the angles, and this can help solve many trigonometric word problems. One of the most common uses of right-angled triangles is the use of the Pythagoras theorem, which is Altitude 2 + Base 2 = Hypotenuse 2 . This allows you to find the measures of missing sides.

When working with right-angled triangles or solving problems that involve them, we can use the properties of sines, cosines, and tangents for determining many different missing sides or angles. Here are some other formulas that can be used with right-angled triangles to identify its unknown parts.

Sines: sin A = a/c, sin B = b/c | Cosines: cos A = b/c, cos B = a/c | Tangents: tan A = a/b, tan B = b/a

Let's just look at some of the cases where we don't know all the sides. Suppose we don't know the sides, but we are familiar with the other two sides. Using the Pythagoras theorem, we can easily identify the remaining two sides. If you know the measure of a side and angle, you can often determine all the others in this geometric environment which makes it a very handy shape to understand.

How Does This Apply to Real Life?

Understanding how to use and manipulate right triangles is one of the most important skills that covered in trigonometry. All of these geometric shapes consist of a single angle that measures ninety degrees. This allows use to use the Pythagorean theorem to our advantage to determine any unknown side or angle within the figure as long as we have a reference side or angle to work off of. In your everyday life you can use this to help achieve very similar things. The minute that you lean a ladder against a straight wall, you have created a right triangle. Want to know the height of the ladder, at what point it touches the wall, or the angle of the ladder, that can all be determined using the same exact math. You can use this math to determine the heights of building or natural structure in much the same way. When we are working with maps to locate areas or plot exact distances between objects, these figures can be pivotal to help you complete this. Anything that involves a coordinate system can be manipulated and better understood with the help of these figures. Mainly because if understand one measure you can learn a great deal about the system it is surrounded in.

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SOLVING RIGHT TRIANGLES WORKSHEET

Problem 1 :

Solve the right triangle shown below and round decimals to the nearest tenth.

Problem 2 :

Problem 3 :

During a space shuttle's approach to earth, it’s glide angle changes.

(i) When the shuttle’s altitude is about 15.7 miles, its horizontal distance to the runway is about 59 miles. What is its glide angle ? Round the answer to the nearest tenth.

(ii)  When the space shuttle is 5 miles from  the runway, its glide angle is about 19°.  Find the shuttle’s altitude at this point in  its descent. Round the answer to the  nearest tenth.

1. Answer :

Begin by using the Pythagorean Theorem to find the length of the hypotenuse.

Pythagorean Theorem : 

(Hypotenuse) 2   =  (Leg) 2 + (Leg) 2

Substitute.

c 2   =  2 2  + 3 2

c 2   =  4 + 9

c 2   =  13

Take square root on each side. 

√ c 2   =   √ 13

c  =   √ 13

Use calculator to approximate.

c  ≈  3.6

Now, use a calculator to find the measure of  ∠B :

Finally, because  ∠A and  ∠B are complements, we can write

m ∠A  =  90 ° -  ∠B   ≈   90 ° - 33.7°  =  56.3°

The side lengths of the triangle are 2, 3, and  √ 13 , or about 3.6. The triangle has one right angle and two acute angles whose measures are about 33.7° and 56.3°.

2. Answer :

Use trigonometric ratios to find the values of g and h.

Because  ∠ H and  ∠ G are complements, we can write

m ∠G  =  90 ° -  ∠H  =   90 ° - 25°  =  65°

The side lengths of the triangle are about 5.5, 11.8, and 13. The triangle has one right angle and two acute angles whose measures are 65° and 25°.

3. Answer :

Sketch a right triangle to model the situation. Let x° be the measure of the shuttle’s glide angle.

We can use the tangent ratio and a calculator to find the approximate value of x.

Write ratio.

tan x °  =  opp. / adj.

tan x °  =  15.7 / 59

x  =   tan -1  (15.7 / 59)

Use calculator to find the value of  tan -1  (15.7 / 59).

x   ≈  14.9

When the space shuttle’s altitude is about 15.7 miles, the glide angle is about 14.9 °.

Part (ii) :

Sketch a right triangle to model the situation. Let h be the altitude of the shuttle. We can use the tangent ratio and a calculator to find the approximate value of h. 

tan 19 °  =  h / 5

Use a calculator.

0.3443    ≈   h / 5

Multiply each side by 5.

5  ⋅  0.3443    ≈   h

1.7    ≈   h

The shuttle’s altitude is about 1.7 miles.

Kindly mail your feedback to   [email protected]

We always appreciate your feedback.

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Right Triangle Trigonometry Worksheets – Full Set (Free Download)

• October 22, 2022
• Math Worksheets

In this article we have covered wide variety of Right Triangle Trigonometry Worksheets that are suitable for middle schoolers. For each section, we have given the methodology that can be used to solve the problems in the worksheet.

Feel free to download and print (for personal use) and try these intuitive trigonometry problems.

Applications of Right Triangle Trigonometry Worksheet

To solve these problems, you will have to first learn the concept of the Pythagorean Theorem and the law of tangents.  

The Pythagorean Theorem  – Is about the relationship between the three sides of a right-angle triangle. So, if ABC is a right-angle triangle in which the three sides are AB, BC, and AC, AB 2 + BC 2 =AC 2 . Here, AC is the hypotenuse-the longest side of the right angle triangle. 

You are needed to use the above formula to determine the unknown side of the right-angled triangle when two sides are already given.

Law of tangents  – It is the relationship between any two sides and the two angles opposite to these two sides of a right-angled triangle ABC. Remember, the law of tangent applies to only right-angle triangles. 

Again tangent of a given angle of a right-angled triangle is the ratio of its opposite side to its adjacent side. 

So, here tanC=AB/BC

Example of finding the length of a side.

Calculate the length of x and y of two right-angle triangles in the figure below that share two vertices and one common straight-line base. Here the length of one side AB is 12, D=60 0 and angle C=30 0 .

To solve the above problem, you have to do the following steps.

Step 1. For triangle ABD, you already have AB=12, B=90 0 and D=60 0

So, tan 60 0 =opposite side/adjacent side

                  =AB/BD

                  =12/BD

                  =12/x

Since it is known from the trigonometry table that tan 60 0 =√3,   you can write

So, x=12/√3

        =3×4/√3

        =√3x√3 x4/√3

        =4√3

Step 2 For triangle ABC, you already have AB=12, B=90 0 and C=30 0

So, tan 30 0 =opposite side/adjacent side

                  =AB/BC

                  =12/(BD +DC)

                  =12/(x+y)

Since it is known from the trigonometry table that tan 30 0 =1/√3,   you can write

1/√3=12/(4√3 +y)

So, (4√3 +y)=12x√3

                 y =12√3-4√3

                    =√3(12-4)

                    =8√3

Now since √3=1.732, you have

       x=4×1.732

         =6.928=7 (approximately )

and y=8×1.732

        =13.85=14 (approximately)

Worksheet for you to try :

Basic Right Triangle Trigonometry Worksheets

To solve these problems, you have to use the law of tangents, sines, and cosines as per the given adjacent side, opposite side and the hypotenuse. 

Example of finding the angle of a right angle triangle when the adjacent side and the hypotenuse are already given

Calculate the angle x of the right-angle triangle in the figure below. Here, the length of one side BC =3 and the hypotenuse AC=6.

Since you have the length of the adjacent side of the angle C and the hypotenuse, you have to use the law of cosine.

So, cos x=adjacent side/hypotenuse

                  =BC/AC

                  =3/6

                  =1/2

Now, as we know cos 60 0 =1/2, so , x=60 0

Example of finding the angle of a right-angle triangle when the opposite side and the hypotenuse are already given

Calculate the angle x of the right-angle triangle in the figure below. Here the length of one side BC =14 and the hypotenuse AC=21.

Since, you have the length of the opposite side of the angle A and the hypotenuse, you have to use the law of sine

So, sin x=opposite side/hypotenuse

                  =14/21

                  =2/3

                 =0.666

Now as sin 41.8 0 =0.666, so, x=41.8 0

Worksheets to try:

Practice Worksheet – Right Triangle Trigonometry

Example of finding sin 45 degrees and cos 45 degrees of the right angle triangle in fraction form when all the three sides are given.

Sin 45 0 =opposite side/hypotenuse

            = 4/7.2

            =40/72

            =10/18

            =5/9

Cos45 0 =adjacent side/hypotenuse

            = 6/7.2

            =60/72

            =10/12

            =5/6

Precalculus Right Triangle Trigonometry Worksheet

Example A swimmer is 210 meter below the surface of the ocean and begins to descend at an angle of 30 degrees from the vertical. How far will be the swimmer travel before he breaks the surface of the water.

You know here Sin 30 0 =opposite side/hypotenuse

           So hyotenuse =210 ÷0.5

                                  =420

So, the swimmer travels 420 meter before he breaks the ocean surface.        

Download              

Right Triangle Trigonometry – Angle of Elevation and Depression Worksheet

To solve the problems in the worksheet, you need to first know the concept of the angle of elevation and the angle of depression.

The angle of elevation -The angle formed when an observer looks at an object above his horizontal line of sight. For example, if you stand on a plateau and look at the peak of a nearby mountain, an angle of elevation is formed.

The angle of depression -The angle formed when an observer looks down at an object below his horizontal line of sight. For example, if you stand on a plateau and look at a house in the plains, an angle of depression is formed.            

Example- A bird is sitting on an iceberg 100 feet above the water. If a sea lion in water is 220 feet from the base of the iceberg, find the angle of depression.

Let x is the angle of elevation that the sea lion at a distance of 220 meter from the base of the iceberg makes when he looks at the bird sitting on the iceberg at a vertical height of 100 meters.

From the above figure, it becomes apparent that tan x 0 =adjacent side /opposite side

                                                                                       =220/100

                                                                                       =11/5

                                                                             tan 66 0 =2.2

Now as tan 66 0 =2.2,you have angle of elevation x= 66 0

Next, we know that the sum of all three angles of a triangle equals to 180 0

So, the remaining angle will be =180 0 –(66 0 + 90 0 ) = 180 0 – 156 0 =24 0

Hence, the angle of depression = 90 0 – 24 0 =66 0

Example A 25 feet ladder leans against a house so that the base of the ladder is 7 feet from the base of the house. What will be the angle of elevation of the ladder.

cos x=adjacent side/hypotenuse

        =7/14

   So, angle of elevation of the ladder= 60 0  

Special Right Triangle Trigonometry Worksheet

Example-Half of an equilateral triangle is often called “30-60” right or “30-60-90” triangle. Explain why it is called with that name?

Since, an equilateral triangle can be split into two right angle triangles with the remaining angles being 30 degrees and 60 degrees, half of an equilateral triangle is often called “30-60” right or “30-60-90” triangle.  Let us prove this with the help of the law of sine and cosine.                     

Let ABC is an equilateral triangle with sides x, y and z where x=y=z.

Now you draw a perpendicular from vertex A on the side BC, you will have two right angle triangles-ABD and ACD with angle D=90 0 .

Next, as per the law of cosine, you know in the right-angle triangle ADC

cos C=adjacent side/hypotenuse

          =DC/AC

          =Half of the side z/y

         =y/2 ÷ y

         =1/2

Now from the trigonometry table, you know that cos 60 0 =1/2

So,      ∠C=60 0

Similarly, as per the law of sine, you know  that in the right-angle triangle ADC

Sin A=opposite side/hypotenuse

         = DC/AC

Now from the trigonometry table, you know that sin30 0 =1/2

So,     ∠A=30 0

Hence, you have in the right-angle triangle ACD, A=30 0 , C=60 0 , and D=90 0

Similarly, in the right-angle triangle ABD, A=30 0 , B=60 0 , and D=90 0

Right Triangle Trigonometry Finding Missing Sides & Angles Worksheet

Right Triangle Trigonometry Word Problems Worksheet

Trigonometry Ratios in Right Triangles Worksheet

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How to Solve Triangles Problems? (+FREE Worksheet!)

Learn how to solve mathematics problems related to triangle area and angles using common triangle formulas.

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A step-by-step guide to solving Triangles

• In any triangle, the sum of all angles is \(180\) degrees.

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Triangles – example 1:.

Use the area formula: Area \(= \color{blue}{\frac{1}{2 }(base \ × \ height)}\) base \(=10\) and height \(=6\) Area \(= \frac{1}{2} (10×6)=\frac{1}{2} (60)=30\)

Triangles – Example 2:

What is the missing angle of the following triangle?

All angles in a triangle sum up to \(180\) degrees. Then: : \(45+60+x=180 → 105+x=180 → x=180-105=75\)

Triangles – Example 3:

What is the area of the following triangle?

Use the area formula: Area \(= \color{blue}{\frac{1}{2 }(base \ × \ height)}\) base \(=12\) and height \(=8\) Area \(= \frac{1}{2} (12×8)= \frac{1}{2} (96)=48 \)

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Triangles – example 4:.

What is the area of a triangle with base \(6\) and height \(5\)?

Use the area formula: Area \(= \color{blue}{\frac{1}{2 }(base \ × \ height)}\) base \(=6\) and height \(=5\) Area \(= \frac{1}{2 } (5×6)= \frac{1}{2} (30)=15 \)

Exercises for Solving Triangles

Find the measure of the unknown angle in each triangle., download triangles worksheet.

• \(\color{blue}{45^{\circ}}\)
• \( \color{blue}{ 15^{\circ}}\)
• \( \color{blue}{ 40^{\circ}}\)

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by: Effortless Math Team about 4 years ago (category: Articles , Free Math Worksheets )

Effortless Math Team

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