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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.

Content Continues Below

## MathHelp.com 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

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:

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!:

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.

• 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 .

URL: https://www.purplemath.com/modules/rghtprob.htm

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## Trigonometry : Solving Word Problems with Trigonometry

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : solving word problems with trigonometry. You can draw the following right triangle using the information given by the question: Since you want to find the height of the platform, you will need to use tangent. You can draw the following right triangle from the information given by the question. In order to find the height of the flagpole, you will need to use tangent. You can draw the following right triangle from the information given in the question: In order to find out how far up the ladder goes, you will need to use sine. ## Example Question #4 : Solving Word Problems With Trigonometry

In right triangle ABC, where angle A measures 90 degrees, side AB measures 15 and side AC measures 36, what is the length of side BC? This triangle cannot exist. ## Example Question #5 : Solving Word Problems With Trigonometry

A support wire is anchored 10 meters up from the base of a flagpole, and the wire makes a 25 o angle with the ground. How long is the wire, w? Round your answer to two decimal places.

23.81 meters 28.31 meters

21.83 meters

To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o , the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w. Now, we just need to solve for w using the information given in the diagram. We need to ask ourselves which parts of a triangle 10 and w are relative to our known angle of 25 o . 10 is opposite this angle, and w is the hypotenuse. Now, ask yourself which trig function(s) relate opposite and hypotenuse. There are two correct options: sine and cosecant. Using sine is probably the most common, but both options are detailed below.

We know that sine of a given angle is equal to the opposite divided by the hypotenuse, and cosecant of an angle is equal to the hypotenuse divided by the opposite (just the reciprocal of the sine function). Therefore: To solve this problem instead using the cosecant function, we would get: The reason that we got 23.7 here and 23.81 above is due to differences in rounding in the middle of the problem. ## Example Question #6 : Solving Word Problems With Trigonometry

When the sun is 22 o above the horizon, how long is the shadow cast by a building that is 60 meters high?

To solve this problem, first set up a diagram that shows all of the info given in the problem. Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we'll be solving for. We'll call this base b. Therefore the shadow cast by the building is 150 meters long.

If you got one of the incorrect answers, you may have used sine or cosine instead of tangent, or you may have used the tangent function but inverted the fraction (adjacent over opposite instead of opposite over adjacent.)

## Example Question #7 : Solving Word Problems With Trigonometry

From the top of a lighthouse that sits 105 meters above the sea, the angle of depression of a boat is 19 o . How far from the boat is the top of the lighthouse?

423.18 meters

318.18 meters

36.15 meters

110.53 meters

To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is. Merging together the given info and this diagram, we know that the angle of depression is 19 o  and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we'll use that here. We get: ## Example Question #8 : Solving Word Problems With Trigonometry

Angelina just got a new car, and she wants to ride it to the top of a mountain and visit a lookout point. If she drives 4000 meters along a road that is inclined 22 o to the horizontal, how high above her starting point is she when she arrives at the lookout?

9.37 meters

1480 meters

3708.74 meters

10677.87 meters

1616.1 meters

As with other trig problems, begin with a sketch of a diagram of the given and sought after information. Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is 22 o . Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled x in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up: Therefore the change in height between Angelina's starting and ending points is 1480 meters.

## Example Question #9 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 50 feet apart. The shorter building is 40 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 48 o . How high is the taller building?

To solve this problem, let's start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle. ## Example Question #10 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 80 feet apart. The shorter building is 55 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 32 o . How high is the taller building? ## Report an issue with this question

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## Unit 1: Right triangles & trigonometry

• Getting ready for right triangles and trigonometry (Opens a modal)
• Hypotenuse, opposite, and adjacent (Opens a modal)
• Side ratios in right triangles as a function of the angles (Opens a modal)
• Using similarity to estimate ratio between side lengths (Opens a modal)
• Using right triangle ratios to approximate angle measure (Opens a modal)
• Right triangles & trigonometry: FAQ (Opens a modal)
• Use ratios in right triangles Get 3 of 4 questions to level up!

## Introduction to the trigonometric ratios

• Triangle similarity & the trigonometric ratios (Opens a modal)
• Trigonometric ratios in right triangles (Opens a modal)
• Trigonometric ratios in right triangles Get 3 of 4 questions to level up!

## Solving for a side in a right triangle using the trigonometric ratios

• Solving for a side in right triangles with trigonometry (Opens a modal)
• Solve for a side in right triangles Get 3 of 4 questions to level up!

## Solving for an angle in a right triangle using the trigonometric ratios

• Intro to inverse trig functions (Opens a modal)
• Solve for an angle in right triangles Get 3 of 4 questions to level up!

## Sine and cosine of complementary angles

• Intro to the Pythagorean trig identity (Opens a modal)
• Sine & cosine of complementary angles (Opens a modal)
• Using complementary angles (Opens a modal)
• Trig word problem: complementary angles (Opens a modal)
• Trig challenge problem: trig values & side ratios (Opens a modal)
• Trig ratios of special triangles (Opens a modal)
• Relate ratios in right triangles Get 3 of 4 questions to level up! ## Modeling with right triangles

• Right triangle word problem (Opens a modal)
• Angles of elevation and depression (Opens a modal)
• Right triangle trigonometry review (Opens a modal)
• Right triangle trigonometry word problems Get 3 of 4 questions to level up!

## The reciprocal trigonometric ratios

• Reciprocal trig ratios (Opens a modal)
• Finding reciprocal trig ratios (Opens a modal)
• Using reciprocal trig ratios (Opens a modal)
• Trigonometric ratios review (Opens a modal)
• Reciprocal trig ratios Get 5 of 7 questions to level up!

## WORD PROBLEMS INVOLVING RIGHT TRIANGLE

Example 1 :

The size of a computer monitor is the length  across its diagonal. If a computer monitor is  34.4 cm long and 27.5 cm high, what size is it? In figure, given

Height of a computer monitor (a)  =  27.5 cm

Length of base (b)  =  34.4 cm

Let c be the length of diagonal.

By using Pythagoras theorem,

a 2 + b 2   =  c 2

(27.5) 2  + (34.4) 2   =  c 2

c 2  =   756.25 + 1183.36

c 2   =  1939.61

c  =  √1939.61

c  =  44.0 cm

So, length of the diagonal c is 44.0 cm.

Example 2 : (a)  Find the length of the wire shown supporting  the TV mast.

(b) There are six wires which support the mast.

i)  Find their total length.

ii)  If 3% extra wire is needed for tying,  how many meters of wire need to be  purchased?

(a)  By finding length of hypotenuse, we can find the length of wire needed.

In figure,

a  =  6.7 m and b  =  7.8 m

To find the length of the wire c  =  ?

(6.7) 2  + (7.8) 2   =  c 2

44.89 + 60.84  =  c 2

105.73  =  c 2

c  =  √105.73

c  =  10.28 m

So, length of the wire c is 10.28 m.

(i)  Length of the one wire is 10.28 m.

Length of 6 wires  =   10.28 × 6

=  61.68 m

So, total length is 61.7 m

(ii) If 3% extra wire is needed for tying, how many meters of wire need t o be purchased? The wire must be purchased in a whole number of meters .

We need 3% extra wire,

=  6 + 3% of 61.68

=  6.18

So, total wires 6.18

Total meters of wires purchased

=  Length of the wire × total wires

=  10.28 × 6.18

=  63.53 m

So, 64 m of wire to be purchased.

Example 3 :

Metal supports are made as shown. They are fitted from the lower edge of the table top to the legs. The flat ends of the supports are 2 cm long. Find the length of the metal needed to make the 8 supports used to stabilise the table. We see this right triangle,

Top metal fits a  =  10 cm

Lower edge metal fits b  =  10 cm

Length of the metal support c  =  ?

10 2  + 10 2   =  c 2

100 + 100  =  c 2

200  =  c 2

c  =  √200

c  =  14.14 cm

Length of the metal supports c  =  14.14 cm

Length of the metal supports 2 cm long top and lower,

So, Length of the metal supports  =  18.14 cm

Length of the metal 8 supports  =  18.14 × 8

=  145.12 cm

Therefore, Length of the metal 8 supports is  145.12 cm.

Example 4 : A ladder is 2 meters long. It leans against a wall  so the base is 86 cm from the wall.

a) Write 86 cm in meters.

b) How far up the wall does the ladder reach?

(a)  86 cm  =  0.86 m

(b)  Using Pythagorean theorem, we get

Let x be the distance reached by the ladder.

(0.86) 2 + x 2   =  2 2

0.7396 +  x 2   =  4

x 2   =  4-0.7396

x 2   =  3.2604

x  =  1.80

Example 5 :

Heidi has made a rectangular cushion cover which  measures 36 cm long by 32 cm wide. She wants to  put lace across one diagonal on each side.

a) Find the length of the diagonal.

b) Find the total length of lace required, if 4 cm  extra is allowed for finishing off the ends.

c) Calculate the cost of the lace at $2.30 per meter (a) Let x be the length of diagonal. Using Pythagorean theorem, 36 2 + 32 2 = x 2 1296 + 1024 = x 2 x 2 = 2320 x = 48.16 x = 48.2 m (b) Total length of lace = 48.2(2) = 96.4 4 m is needed, then length of lace = 96.4+4 = 100.4 (c) Required cost = 100.4(2.30) =$231

Example 6 :

A square garden with sides 100 m is divided into two triangular plots by a fence along one diagonal.

a) What is the length of the fence in meters (to 1 decimal place) ?

b) If the fence costs $15.50 per meter, what is the total cost ? Let x be the length of diagonal. x 2 = 100 2 +100 2 x 2 = 10000+10000 x 2 = 20000 x = 141.4 m (b) Cost of fencing =$5.50 per meter

Required cost  =  141.4(5.50)

=  $777.7 Example 7 : A 160 m long water pipe runs along the diagonal of a square paddock. What are the lengths of the sides of the paddock? Length of diagonal pipe = 160 m Let x be the side length square paddock. x 2 +x 2 = 160 2 2x 2 = 25600 x = 113.14 m So, side length of paddock is 113.14 m Example 8 : A garden gate is 1.2 meters wide and 1.2 meters high. The gate is strengthened by a diagonal strut. a) How long is the strut? b) Calculate the length of steel needed for the frame of the gate, including the strut. x 2 = (1.2) 2 + (1.2) 2 x 2 = 1.44+1.44 x 2 = 2.88 x = 1.7 m (b) Length of steel needed including the diagonal = 4(1.2) + 1.7 = 4.8+1.7 = 6.5 m Kindly mail your feedback to [email protected] We always appreciate your feedback. © All rights reserved. onlinemath4all.com • Sat Math Practice • SAT Math Worksheets • PEMDAS Rule • BODMAS rule • GEMDAS Order of Operations • Math Calculators • Transformations of Functions • Order of rotational symmetry • Lines of symmetry • Compound Angles • Quantitative Aptitude Tricks • Trigonometric ratio table • Word Problems • Times Table Shortcuts • 10th CBSE solution • PSAT Math Preparation • Privacy Policy • Laws of Exponents ## Recent Articles ## Equality of Matrices Nov 02, 23 09:37 PM ## Eigenvalues of a Matrix Oct 30, 23 09:23 PM ## Laws of Exponents Worksheet Oct 30, 23 09:09 PM • Daily Games • Strategy and Puzzles • Vocabulary Games • Junior Edition Games • All problems • High School Math • MAML Problems • Calculus Problems • Loony Physics • Pro Problems • Getting Started • Pro Control Panel • Virtual Classroom • Play My Favorites • Select My Favorites ## Trig Word Problems #1 Now that we have a basic understanding of what the trig functions sine, cosine, and tangent represent, and we can use our calculators to find values of trig functions, we can use all of this to solve some word problems. In this reading we'll simply look at examples of word problems, and then let you give them a try. Sample #1 The sun's angle of inclination is 20 degrees, and a pole casts a 40 foot shadow. How tall is the pole? Solution Using the image above, X = 20 degrees, and y = 40 ft. tan X = x / y 0.3640 = x / 40 x = 14.56 ft Sample #2 A ramp is 50 feet long, and it is set at a 30 degree angle of inclination. If you walk up the ramp, how high off the ground will you be? Solution Using the image above, X = 30 degrees and z = 50 ft. sin X = x / z 0.5 = x / 50 x = 25 Sample #3 A man walks 5 miles at 60 degrees north of east. How far east of his starting point is he? Solution Using the image above, with y representing the eastern travel, x representing the northern travel, and z representing the actual path of the man, sin X = x / z 0.8660 = x / 5 x=4.33 miles. ## Blogs on This Site ## 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? Mometrix Academy – Home by Mometrix Test Preparation | This Page Last Updated: June 27, 2023 ## Solving Right Triangles Right triangles are geometric figures that offer a variety of applications. It is widely used for construction purposes, navigation, approximation of measurement of tall structures, and a lot more. Furthermore, it allowed us to develop trigonometry , extensively used in other fields such as astronomy , physics , and engineering. How can such a simple figure become so powerful that it changes the course of our planet? Learn more about this geometric plane figure and techniques for solving right triangles in this review. Click below to go to the main reviewers: Ultimate PMA Entrance Exam Reviewer Ultimate UPCAT Reviewer Ultimate LET Reviewer ## Table of Contents What are right triangles: a review. Recall that we can classify triangles according to interior angles . If the interior angles of a triangle are all acute, then we call it an acute triangle . If at least one interior angle is obtuse, we call it an obtuse triangle . Now, if a triangle has one 90° angle, it is a right triangle. Essentially, right triangles contain a right-angle (90° angle) while the two interior angles are acute. ## Parts of a Right Triangle The longest side of a right triangle is called the hypotenuse. Meanwhile, the remaining shorter sides of the right triangle are called the legs. To help you remember: • Longest side ⇰ Hypotenuse • Shorter sides ⇰ Legs The small square that you can see between the legs of the right triangle indicates that the angle formed by the legs is a 90-degree angle. Once you see this “tiny square” inside a triangle, it gives you a hint that the triangle you are looking at is a right triangle. In the figure above, sides PQ and QR are the legs of the right triangle PQR, while side PR is the hypotenuse. Take note that the hypotenuse of a right triangle is always opposite to the right angle. In a right triangle, the legs are perpendicular since they form a right angle. Hence, in the given figure above PQ ⊥ QR. If the legs of the right triangle are congruent or equal in measurement, then that right triangle is called an isosceles right triangle . Meanwhile, if all sides of the right triangle have different measurements, it is a scalene right triangle . Now that you know a right triangle’s legs and the hypotenuse, let us explore the theorem that shows the relationship among these parts. ## Pythagorean Theorem “The sum of the squares of the lengths of the legs of a right triangle on a flat surface is equal to the square of the length of its hypotenuse.” The Pythagorean theorem states that if you square the lengths of both legs of the right triangle and then add them, the result will be equivalent to the square of the length of the hypotenuse. (Leg 1) 2 + (Leg 2) 2 = (Hypotenuse) 2 To make the mathematical statement above easier to remember, we let a and b be the length of the legs and c be the hypotenuse of the right triangle. The Pythagorean theorem states that: a 2 + b 2 = c 2 This theorem is one of the most popular theorems in mathematics. It’s attributed to the Greek mathematician Pythagoras who proved this theorem hundreds of years ago. If you are interested in how this theorem has been proven throughout the years, you may watch this video from Ted-Ed. Sample Problem 1: The right triangle below has legs measuring 4 cm and 3 cm. Determine how long its hypotenuse (or the longest side) is. Solution: The Pythagorean theorem states that a 2 + b 2 = c 2 where a and b are the legs while c is the hypotenuse. We have a = 4 and b = 3. Using the Pythagorean theorem, let us solve for the hypotenuse c : a 2 + b 2 = c 2 (4) 2 + (3) 2 = c 2 Substitute a = 4 and b = 3 16 + 9 = c 2 √25 = √c² Take the square root of both sides of the equation The values of c that we have obtained are 5 and -5. Recall that c represents the length of the hypotenuse of the right triangle. Thus, we must reject -5 since having a negative length of a right triangle is impossible. Thus, we will only take 5 as the value of c. Hence, the hypotenuse of the right triangle is 5 cm. Sample Problem 2: Look at the map shown below. How long is the shortest path from Helen’s house to the library? Solution: The shortest path connecting Helen’s house and the library is the diagonal line connecting these two places. So, we draw a diagonal line that connects them. The length of the diagonal line that we draw is the distance of the shortest path from Helen’s house to the library. Notice that we have created a right triangle as we draw this diagonal line between Helen’s house and the library. The legs are the road from Helen’s house to the Acacia tree and the road from the Acacia tree to the library. And what is the hypotenuse? None other than the diagonal line representing the shortest path from Helen’s house to the library. Since we have formed a right triangle, we can apply the Pythagorean theorem to find the length of this diagonal line. We have already defined the legs of this triangle in the previous paragraph. The diagonal line will serve as the hypotenuse of this right triangle. Let us do the math now. The legs have measurements of 30 m and 40 m. So, we have a = 30 and b = 40. We let c as the length of the hypotenuse of the right triangle. Using the Pythagorean theorem: (30) 2 + (40) 2 = c 2 900 + 1600 = c 2 √c² = √2500 We reject the negative value of c. Thus, the hypotenuse of the right triangle we formed is 50. It means that the diagonal line we draw measures 50 m. Therefore, we conclude that the shortest path from Helen’s house to the library is 50 meters long (try computing the total distance of all possible paths from Helen’s house to the library, and you will discover that the shortest distance is indeed the one represented by the diagonal line). Sample Problem 3: What is the measurement of the side AB in the right triangle below? Solution: We have legs AC and BC that measure 2 and 3 units, respectively. Let us compute the hypotenuse AB using the Pythagorean theorem. a 2 + b 2 = c 2 The a and b are the legs, and c is the hypotenuse (2) 2 + (3) 2 = c 2 Substitute the values or measurement of the legs to a and b 4 + 9 = c 2 c 2 = 13 Symmetric property of equality We reject the negative value. Thus, the hypotenuse of the right triangle is √13 units. Sample Problem 4: Determine the value of Q in the right triangle below. Solution: In the figure above, Q represents the measurement of a leg. The other leg of the right triangle measures 6 cm, while its hypotenuse is 10 cm. To find the missing measurement of a leg of a right triangle, we can use the Pythagorean theorem and manipulate the equation we will form. a 2 + b 2 = c 2 where a and b are the legs and c is the hypotenuse (6) 2 + (Q) 2 = (10) 2 One of the legs is 6 cm while the hypotenuse is 10 cm 36 + Q 2 = 100 Q 2 = -36 + 100 Transposition method (we isolate b from the constants) We reject -8 and take Q = 8 only. Therefore, the missing measurement of the leg of the right triangle is 8 cm long. And since Q represents the measurement of a leg of the right triangle, the value of Q must be 8 cm. ## Pythagorean Triples Pythagorean triples are three positive whole numbers that satisfy the Pythagorean theorem. The smallest Pythagorean triples are 3, 4, and 5. Notice how they satisfy the Pythagorean theorem: If we let a = 3, b = 4, and c = 5: (3) 2 + (4) 2 = (5) 2 9 + 16 = 25 Both sides are equal; therefore, 3, 4, and 5 are Pythagorean triples. Now, let’s try 1, 2, and 3. Are these numbers Pythagorean triples? Let a = 1, b = 2, and c = 3 (1) 2 + (2) 2 = (3) 2 1, 2, and 3 failed to satisfy the Pythagorean theorem, so they are not triples. Some common Pythagorean triples are the following: • 3, 4, and 5 • 5, 12, and 13 • 7, 24, and 25 • 8, 15, and 17 • 11, 60, and 61 You can try it and confirm that these triples are all Pythagorean triples. How can we get Pythagorean triples? If you take the multiples of the triples listed above, you will still obtain a Pythagorean triple. For instance, 5, 12, and 13 is a Pythagorean triple. If we multiply these three numbers by 4, we have 20, 48, and 52; these new triples are also Pythagorean triples. Sample Problem: Is 27, 36, and 45 a Pythagorean triple? Solution: Yes, because 27, 36, and 45 are derived by multiplying each of 3, 4, and 5, which are Pythagorean triples, by 9. ## Converse of the Pythagorean Theorem Again, the Pythagorean theorem tells us that in a right triangle, the sum of the squares of the lengths of legs is equal to the square of the length of the hypotenuse: a 2 + b 2 = c 2 The converse of the Pythagorean theorem tells us that if the square of the length of the longest side of the triangle is equal to the sum of the squares of the lengths of the remaining shorter sides, then the triangle formed is a right triangle . Once we show that the condition above was satisfied, the triangle formed is a right triangle. What if, for instance, the conditions of the converse of the Pythagorean theorem are unsatisfied? We have two possible cases: Case 1: If the square of the longest side is greater than the sum of the squares of the shorter sides of the triangle or c 2 > a 2 + b 2 (where a and b are shorter sides and c is the longest side). In this case, we have an obtuse triangle . Case 2 : If the square of the triangle’s longest side is less than the sum of the squares of the shorter sides or c 2 < a 2 + b 2 , then we have an acute triangle. So, to summarize: If a and b are the shorter sides of a triangle and c is its longest side: • If c 2 = a 2 + b 2 , then we have a right triangle • If c 2 > a 2 + b 2 , then we have an obtuse triangle • If c 2 < a 2 + b 2 , then we have an acute triangle Sample Problem 1 : The shorter sides of a triangle have measurements of 2 cm, 4 cm, and 5 cm. Determine what type of triangle this is. Solution: The square of the length of the longest side (the one that measures 5 cm) is (5) 2 = 25. The sum of the squares of the lengths of the shorter sides (the remaining sides) is: (2) 2 + (4) 2 = 4 + 16 = 20 Notice that the square of the length of the longest side is greater than the sum of the squares of the lengths of the shorter sides or c 2 > a 2 + b 2 . Thus, the triangle is obtuse. Sample Problem 2: Triangle PQR has the following measurements: PQ = 5 cm, QR = 2 cm, and PR = 7 cm. Determine whether the triangle is acute, right, or obtuse. Solution: The triangle’s longest side is PR, which measures 7 cm. Its square is (7) 2 = 49. The sum of the squares of the lengths of the shorter sides PQ and QR is: (5) 2 + (2) 2 = 25 + 4 = 29 Notice that the square of the length of the longest side is greater than the sum of the squares of the lengths of the shorter sides or c 2 > a 2 + b 2 . Thus, triangle PQR is an obtuse triangle. ## Special Right Triangles In the previous sections, you have learned how special right triangles are because of their wonderful characteristics described by the Pythagorean theorem. But did you know that there are special right triangles? This special plane figure has some types that are considered more special! There are two types of special right triangles : the isosceles right triangles (or the 45° – 45° – 90° right triangle) and the 30° – 60° – 90° degree right triangles. Let us talk about these special right triangles in this section. ## 1. Isosceles Right Triangles (45° – 45° – 90° Right Triangle) An isosceles right triangle is a right triangle such that its legs are congruent or have the same measurement. In an isosceles right triangle, the acute interior angles near the hypotenuse have degree measurements of 45°. This is why we call isosceles right triangles 45° – 45° – 90° right triangles. What makes isosceles right triangles special? In an isosceles right triangle, the length of the hypotenuse is √2 times longer than the measurement of a leg. Thus, you can determine the hypotenuse of the right triangle simply by multiplying the measurement of a leg by √2. No need to use the Pythagorean theorem! Can you imagine how quick and easy it is? We formally state this property above as a theorem: ## Isosceles Right Triangle Theorem “If a right triangle is an isosceles right triangle (or 45°- 45°- 90° right triangle), then the hypotenuse is √ 2 times as long as the leg.” Sample Problem 1: Determine the measurement of the hypotenuse of the right triangle below. Solution: Since the given right triangle is an isosceles right triangle, we can multiply the leg measurement by √2 to obtain the hypotenuse. Thus, if the measurement of the leg is 5 cm, then the hypotenuse is simply 5√2 cm long. Sample Problem 2: Determine how long a square’s diagonal is with a side measuring 2 meters. Let us draw this square and its diagonal. Solution: As we draw the square’s diagonal, we have formed a right triangle such that the legs are the sides and the hypotenuse is the diagonal. Interestingly, the right triangle formed is an isosceles right triangle since the legs are 2 meters long. Thus, the hypotenuse of this isosceles right triangle we form is 2√2 meters. Since the hypotenuse of this isosceles right triangle is also the square’s diagonal, the diagonal is 2√2 meters long. In the previous example, we have derived a practical way to measure a square’s diagonal, given only the square’s side. The square’s diagonal is simply the measurement of the side times √2. This fantastic concept was a direct consequence of the isosceles right triangle theorem. Diagonal of a square = s√2 where s is the measurement of a side. Sample Problem 3: The right triangle below has a hypotenuse with a length of 10 cm. Determine the length of a leg of the right triangle. Answer: Although the right triangle above seems peculiar from the right triangles we have encountered earlier, it is still a right triangle because it has a right angle. Anyway, the right triangle above is an isosceles right triangle because of the two 45-degree angles you can see near the hypotenuse with a length of 10 cm. How do we find the measurement of a leg? According to the isosceles right triangle theorem, the hypotenuse is √2 times as long as the leg. Thus, the measurement of a leg can be derived if we divide the measurement of the hypotenuse by √2. Thus, the measurement of the leg of the isosceles right triangle below is again just 10 divided by √2 : Measurement of the leg = 10 ÷ √2 We can compute the measurement of the leg as follows: 10 ÷ √2 = 10⁄√2 Notice that we have a radical in the denominator, so we need to rationalize it by multiplying both the numerator and the denominator by √2: Thus, the leg measures 5√2 cm. ## 2. 30° – 60° – 90° Right Triangle A 30° – 60° – 90° right triangle is another type of special right triangle. Unlike the previous special right triangle we have discussed, the 30° – 60° – 90° right triangle has no congruent sides. Furthermore, its acute interior angles have different measurements. One is a 30° angle, while the other is a 60° angle. The side of the 30° – 60° – 90° right triangle that is opposite the 30° angle is the shorter leg. Meanwhile, the side opposite the 60° angle is the longer leg. Of course, the longest side is still the hypotenuse opposite the right angle. • The side opposite the 30-degree angle = shorter leg • The side opposite the 60-degree angle = longer leg • Longest side = hypotenuse ## 30° – 60° – 90° Right Triangle Theorem “In a 30° – 60° – 90° right triangle, the hypotenuse is twice as long as the shorter leg. Meanwhile, the longer leg is √ 3 times longer than the shorter side.” In this kind of special right triangle, the hypotenuse measurement can be obtained by simply doubling (or multiplying by 2) the length of the shorter side. Meanwhile, the measurement of the longer leg can be obtained by multiplying the length of the shorter leg by √3. In symbols, if x is the measurement of the shorter leg, then: • Longer leg = x√3 • Hypotenuse = 2x Look at the 30° – 60° – 90° right triangle below. Its shorter leg measures 2 cm. The longer leg is 2√3 cm. Notice that the longer leg length is just the measure of the shorter leg times √3. Meanwhile, its hypotenuse is 4 cm long, just twice the shorter leg’s measure. Sample Problem 1: Determine the x and y values in the right triangle below. Solution: Obviously, the given right triangle above is a 30° – 60° – 90° right triangle because of the existence of 30-degree and 60-degree angles inside it. To make our computation easy, let us determine first the shorter leg. Recall that the shorter leg is the leg that is opposite to the 30-degree angle. If you take a look at the given image, the shorter leg is the one that has a measure of 5 cm. Meanwhile, the longer leg is the side with the letter y as the measurement since this side is opposite the 60-degree angle. Lastly, x represents the right triangle’s hypotenuse or the longest side. So, we have the following given: • Shorter leg = 5 cm • Longer leg = y • Hypotenuse = x In a 30° – 60° – 90° right triangle, the hypotenuse length is just twice the length of the shorter leg. Since the shorter leg is 5 cm, the hypotenuse is just 5(2) = 10 cm long. Since x represents the hypotenuse of the right triangle, then x = 10 cm. Meanwhile, the length of the longer leg is √3 times as long as the length of the shorter leg. Thus, if the shorter leg is 5 cm long, the longer leg is 5(√3) = 5√3 cm long. Since y represents the longer leg, then y = 5√3 cm. Therefore, x = 10 cm while y = 5√3 cm. Sample Problem 2: Determine the measurement of side p in the figure below: Solution: First, we must identify what p represents in the figure above. The figure above is a 30° – 60° – 90° right triangle. The side with the letter p is the shorter leg of this 30° – 60° – 90° right triangle because this side is opposite the 30-degree angle. Meanwhile, the side with a length of 8√3 is the longer leg since this side is opposite the 60-degree angle. Recall that in a 30° – 60° – 90° right triangle, the longer leg is √3 times longer than the shorter leg. Now that we have 8√3 as the longer leg, how do we derive the length of the shorter leg? Yes, we have to divide 8√3 by √3: 8√3 ÷ √3 = 8 Thus, the shorter leg is 8 cm long. The measurement of the shorter leg is represented by p ; therefore, p = 8 cm. ## Solving Word Problems Involving Right Triangles It’s now time to apply what you have learned about right triangles. Let us try to solve some word problems involving right triangles by using the Pythagorean theorem and the characteristics of the special right triangles. Sample Problem 1: The diagonal of a rectangle is 35 cm long. If the width of the rectangle is 21 cm long, determine the perimeter of the rectangle. Solution: The problem above can be solved easily if we create a diagram to illustrate it. So, we have a diagonal that is 35 cm long and a width of 21 cm long: The problem requires us to find the perimeter of the rectangle above. Recall that the formula for the perimeter of a rectangle is P = 2l + 2w where l is the length and w is the rectangle’s width. We already have the width of the rectangle. Unfortunately, the problem didn’t provide us with the length. So, our goal is to look for the length of the rectangle, and then once we find it, we will compute the perimeter of the rectangle. How can we find the length of the rectangle using the given values in the problem? If you look again at the given rectangle above, the diagonal formed a right triangle with the diagonal as the hypotenuse and the width as one of the legs of the right triangle. This implies that the other leg of this right triangle must be the length of the rectangle. Since the length of the rectangle is unknown, we let x be the length of the rectangle. We can now determine the length of the rectangle using the Pythagorean theorem. The Pythagorean theorem states that a 2 + b 2 = c 2 where a and b are the legs of the right triangle and c is the hypotenuse. Since our legs are the length and the width of the rectangle, then we have the following: • a = x (this is the value of our unknown length) • b = 21 (this is the given width of the rectangle) Furthermore, since the rectangle’s diagonal is the hypotenuse, we have c = 35. Let us now do the math and compute for the length (which is x): (x) 2 + (21) 2 = (35) 2 Using the values we have set above x 2 + 441 = 1,225 x 2 = -441 + 1,225 Transposition method √x² = √784 Take the square root of both sides We will disregard the negative value of x and take x = 28 only. Since x represents the length of the rectangle, then the length of the rectangle is 28 cm. We can now determine the perimeter of the rectangle since we have the length and width measurements. We have length = 28 cm and width = 21 cm, so: P = 2l + 2w P = 2(28) + 2(21) P = 56 + 42 Thus, the perimeter of the rectangle is 98 cm. Sample Problem 2: A ladder measuring 5 meters long is leaning against a brick wall. The ladder makes a 60° angle from the ground. Determine how tall the brick wall is. Solution: Let us try to illustrate this problem first. So, we have a 5-meter ladder leaning against a wall. That ladder makes a 60-degree angle from the ground. Our task is to find the height of the brick wall. Looking closely at our illustration above, we formed a right triangle such that the ground and the wall are the legs while the ladder is the hypotenuse. Furthermore, since the ladder created a 60-degree angle, it is a 30°-60°-90° right triangle. Hence, the other acute angle in the figure must be a 30 ° angle. Let us determine this right triangle’s shorter and longer leg. The shorter leg is the side that is opposite the 30 ° angle. Thus, the shorter leg is the ground between the ladder and the wall. Meanwhile, the longer leg is the side opposite the 60° angle. Thus, the longer leg is the wall or the height of the wall. Meanwhile, the ladder is the hypotenuse of this right triangle. Thus, to find the height of the wall, we need to find the longer leg of this right triangle. We can start by determining the shorter leg first. According to our previous theorem about 30°- 60°- 90° triangles, the hypotenuse is twice as long as the shorter leg. Since the hypotenuse of the right triangle in our illustration is the ladder, which is 5 meters long, the shorter leg is simply 5⁄2 or 2.5 meters long. Therefore, the shorter leg is 2.5 meters long. The longer leg is √3 times as long as the shorter leg. This means that the measurement of the longer leg is just the product of √3 and the measurement of the shorter leg. Since we have computed that the shorter leg’s measurement is 2.5 meters long, the longer leg’s measurement must be 2.5√3 meters. Recall that the height of the wall in the problem is equivalent to the longer leg of the right triangle formed. Thus, the height of the wall in the problem is 2.5√3 meters. Sample Problem 3: A right triangle is inscribed in a circle. Suppose the legs of the right triangle are both 12 cm long. Determine the radius of the circle. Solution: Based on our previous discussion about circles, the right triangle’s hypotenuse will be the circle’s diameter when a right triangle is inscribed in a circle . Thus, to find the circle’s radius in this problem, we can solve for the hypotenuse of the right triangle first since it is also the circle’s diameter. Afterward, we will divide the result by 2 to obtain the radius (since the circle’s diameter equals twice the circle’s radius). The given legs of the right triangle are both 12 cm. Hence, we can conclude that this right triangle is an isosceles right triangle. According to the isosceles right triangle theorem, the length of the hypotenuse of an isosceles right triangle is √2 times as long as the length of the legs. Thus, if the legs of the right triangle in the problem are 12 cm long, the hypotenuse must be 12√2 cm long. Again, since the hypotenuse is also the diameter of the circle, then the diameter of the circle is 12√2 cm. To obtain the radius, we divide 12√2 by 2: 12√2 ÷ 2 = 6√2 Thus, the radius of the circle is 6√2 cm long. Remember the concepts you have learned about right triangles in this review because these will be your primary tools to understand the trigonometric functions we will discuss in the next review. Next topic: Six Trigonometric Functions Previous topic: Circles Return to the main article: The Ultimate Basic Math Reviewer ## Download Printable Summary/Review Notes Download printable flashcards, test yourself, 1. practice questions [free pdf download], 2. answer key [free pdf download], 3. math mock exam + answer key. Jewel Kyle Fabula Jewel Kyle Fabula is a Bachelor of Science in Economics student at the University of the Philippines Diliman. His passion for learning mathematics developed as he competed in some mathematics competitions during his Junior High School years. He loves cats, playing video games, and listening to music. ## Read more Math reviewers: The Ultimate Basic Math Reviewer This basic math reviewer will serve as your key to understand mathematics and discover its practical uses. Circle Problems With Solutions In this chapter, let us view circles from a “geometric lens” and exercise #### IMAGES 1. Right Triangle Word Problems (multiply) 2. Right Triangle Word Problems Worksheet 3. Solving Word Problems Involving Right Triangles 4. How to Solve a Right Triangle in a Word Problem 5. Question Video: Using Right Triangle Trigonometry to Solve Word 6. Day 10 CW Trig Function Word Problems, Solving Right Triangles #### COMMENTS 1. Solving right-triangle word problems: Learn here! 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. MathHelp.com Right Triangle Word Problems 2. Right triangle word problems c 2 = a 2 + b 2 Calculating missing side lengths in right triangles With the Pythagorean theorem, we can calculate any side length in a right triangle when given the other two. Let's look at some examples! 4 6 A B C What is the length of B C ― in the figure above? [Show me!] 25 24 D E F What is the length of E F ― in the figure above? [Show me!] 3. How to Solve a Right Triangle in a Word Problem Step 1: Draw out a simple graphic to represent the word problem, making sure to include a right triangle. Step 2: Label the sides and/or angles of the right triangle that were given in... 4. PDF Right Triangle Word Problems Right Triangle Word Problems 1. Read the problem carefully. 2. Draw and label the triangle. 3. Set up the equation. 4. Solve the equation. 5. Write a therefore statement. 1 A boy who is flying a kite lets out 300 feet of string which makes an angle of 60o with the ground. 5. Trigonometry SOLUTION: • Think of this problem as working with two separate triangles: (1) the larger triangle with the 40º angle and a vertical side that represents the ENTIRE height, b, of the tower, and (2) the smaller triangle with the 25º angle and a vertical side, a, that represents the height of the first (bottom) section of the tower. • Solve for the vertical heights (b and a) in the two ... 6. General triangle word problems (practice) General triangle word problems. The dwarf queen wants to build a tunnel between her cities Yram and Haras. To plan the tunnel's length, the queen first walks 7.0 km from Yram to a point where she can see both cities. From that point, she measures 29 ° between the cities, as shown. Lastly, she walks 6.0 km to Haras. 7. How to Find a Length in a Word Problem with 2 Right Triangles Using Step 1: Let's draw a figure that represents the scenario. Step 2: We will need to find height 1 for the red triangle, which is the side opposite to our angle. Step 3: The problem already gives... 8. Right triangle trigonometry word problems (practice) A dashed line connects from the point to the surface above the location to the left of the point. This forms the hypotenuse of a right triangle that is eighty meters in distance. The location to the left of the point up to the surface is the height of the right triangle. The angle opposite the height is unknown. 9. 10.1: Non-right Triangles 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. 10. Solving Word Problems with Trigonometry Explanation: . To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o, the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w.. Now, we just need to solve for w using the information given in the diagram. 11. PDF Right Triangle Trigonometry: Solving Word Problems In word problems, the formulas remain the same: Word problems introduce two new vocabulary terms: Angle of Elevation The angle of elevation is always measured from the ground up. Think of it like an elevator that only goes up. It is always INSIDE the triangle. In the diagram at the left, x marks the angle of elevation of the top of 12. Right triangles & trigonometry Cue sine, cosine, and tangent, which will help you solve for any side or any angle of a right triangle. Ratios in right triangles Learn Getting ready for right triangles and trigonometry Hypotenuse, opposite, and adjacent Side ratios in right triangles as a function of the angles Using similarity to estimate ratio between side lengths 13. How to Solve Right Triangle Trig Functions Word Problems 296 43K views 8 years ago SAT Math Practice Join us on this lesson where you will get extra practice with modeling and solving right triangle trig functions word problems (right triangle... 14. Word Problems Involving Right Triangle a) Find the length of the diagonal. b) Find the total length of lace required, if 4 cm extra is allowed for finishing off the ends. c) Calculate the cost of the lace at$2.30 per meter. Solution : (a) Let x be the length of diagonal. Using Pythagorean theorem, 362 + 322 = x2.

15. Trig Word Problems #1: Trigonometry

In this reading we'll simply look at examples of word problems, and then let you give them a try. Sample #1. The sun's angle of inclination is 20 degrees, and a pole casts a 40 foot shadow. How tall is the pole? Solution. Using the image above, X = 20 degrees, and y = 40 ft. tan X = x / y. 0.3640 = x / 40. x = 14.56 ft.

16. Master Solving word problems using right triangle trigonometry

Master Solving word problems using right triangle trigonometry Brian McLogan 1.3M subscribers Join Subscribe Share Save 26K views 5 years ago Trigonometric Functions #Master Subscribe!...

17. PDF Right Triangles Day 2

More Word Problems!!!More Word Problems!!! 7. A wall is supported by a brace 10 feet long, as shown in the diagram. If one end of the brace is placed 6 feet from the base of the wall, how many feet up the wall does the brace reach? 8. How many feet from the base of a house must a 39-foot ladder be placed so that the top of the ladder

18. Trigonometry Word Problems

Solution: We can find the height of the monument by using the tangent ratio and then adding the eye height of the student. tan 87.4 ∘ = h 25 h = 25 ⋅ tan 87.4 ∘ = 550.54. Adding 5 ft, the total height of the Washington Monument is 555.54 ft. According to Wikipedia, the actual height of the monument is 555.427 ft.

19. 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 Transcript FAQs Right triangle word problem.

20. Exam Review Solve a word problem using right triangles

👉 Learn how to solve the word problems with trigonometry. Word problems involving angles, including but not limited to: bearings, angle of elevations and de...

21. 4.1.7: Trigonometry Word Problems

Solution. Draw a picture. The angle that the sun hits the flagpole is x∘ x ∘. We need to use the inverse tangent ratio. tan x tan−1 42 25 = 42 25 ≈ 59.2∘ = x tan x = 42 25 tan − 1 42 25 ≈ 59.2 ∘ = x. Example 4.1.7.3 4.1.7. 3. Elise is standing on top of a 50 foot building and sees her friend, Molly.

22. Solve Right Triangles 4 word problem

Using Trigonometry to solve a word problem including right triangles and the angle of elevation when at least one side of the triangle is given and either another side or one of the acute...

23. Solving Right Triangles

Solving Word Problems Involving Right Triangles. It's now time to apply what you have learned about right triangles. Let us try to solve some word problems involving right triangles by using the Pythagorean theorem and the characteristics of the special right triangles. Sample Problem 1: The diagonal of a rectangle is 35 cm long. If the width ...