how to solve word problems with right triangles

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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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High school geometry

Course: high school geometry   >   unit 5.

• Right triangle word problem
• Angles of elevation and depression

Right triangle trigonometry word problems

• Right triangle trigonometry review
• Right triangles and trigonometry FAQ
• 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 ‍  

Right Triangle Word Problems

Exercise 10, exercise 11, exercise 12, solution of exercise 1, solution of exercise 2, solution of exercise 3, solution of exercise 4, solution of exercise 5, solution of exercise 6, solution of exercise 7, solution of exercise 8, solution of exercise 9, solution of exercise 10, solution of exercise 11, solution of exercise 12.

Putting the value of x in equation 1:

Radius of the inscribed circle.

Radius of the circumscribed circle.

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Trigonometric Values

Trigonometric identities, trigonometric ratios, trigonometric functions, trigonometric formulas, systems of simultaneous trigonometric equations, unit circle, sine and cosine law, solving right triangles, even-odd identities, oblique triangles, pythagorean identities, measurement of angles, other trigonometric identities, trigonometric equations, cofunction identities, trigonometric identities problems, oblique triangle word problems, trigonometric angles worksheet, trigonometric equations worksheet, cancel reply.

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Word Problems Using Right Triangles

This video shows how to solve the following word problem involving right triangles: A ramp with a 30° angle from the ground is to be built up to a 2 ft. high platform. How far from the platform will it extend, to the nearest inch?

Right triangle word problem. A ramp with a 30 degree angle from the ground is to be built up to a 2 foot high platform. What will be the length of the ramp? How far from the platform will it extend to the nearest inch? With problems like these, it’s very helpful to draw a picture.

We have our ramp that makes a 30 degree angle with the ground. Let’s start with that. Here’s the ground. Here’s are 30 degree angle. It’s to be built up to a 2 foot high platform. Then here we have a platform, and it’s 2 feet high above the ground. There’s a right angle there perpendicular to the ground.

The first question is; what will be the length of the ramp? This is what we’re trying to find. Well before I start to find X I first want to fill in what this angle is. You might already know that since we have a 30 degree angle and a 90 degree angle that this angle has to be 60 degrees.

But, if you didn’t know that you could find it, because there are 180 degrees in a triangle. If you add 30 to 90 you get 120. 180 minus 120 leaves you with 60 degrees for that third angle. Now that we know it’s a 30 60 90 triangle, we can apply our 30 60 90 rules to finding the length of our ramp.

Our ramp is across from the 90 degree angle, therefore that’s our hypotenuse. The side we know, the leg we know is across from the 30 degree angle, so it’s our shorter leg. The hypotenuse of a 30 60 90 triangle is twice as long as the shorter leg.

Our hypotenuse again, we’re using X for, equals 2 times the shorter leg which is that 2 foot side. 2 times 2 feet. Which means X is 4 feet. Therefore, the length of the ramp is 4 feet. We’ve answered the first question. Then they asked; how far from the platform will it extend, how far from the platform will the ramp extend?

Now they’re asking us to find that length. Y is across from our 60 degree angle so that’s our longer leg. The longer leg of a 30 60 90 triangle is the square root of three times the shorter leg. Our longer leg again is Y equals the square root of three times, our shorter leg is still that two foot length.

Y is 2 square roots of 3 feet. They didn’t ask us for the answer in feet they asked to the nearest inch. The first thing I’m going to do is convert 2 square roots of 3 feet into inches by multiplying by 12. Y is 2 square roots of 3 times 12, since there are 12 inches in one foot, which is 24 square roots of 3 inches.

They didn’t ask for it in simplest radical form, so then we’d actually want to calculate what 24 square root three inches is, and it’s about 42 inches. Therefore, how far from the platform will the ramp extend? The ramp will extend about 42 inches from the platform. There we have it.

Frequently Asked Questions

What is a right triangle.

How do you find the angle of a right triangle?

How do you find the area of a right triangle?

How do you find the sides of a right triangle?

How do you find the hypotenuse of a right triangle?

How do you find the missing side of a right triangle?

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by Mometrix Test Preparation | This Page Last Updated: June 27, 2023