word problems trigonometry with solutions

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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?

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4.1.7: Trigonometry Word Problems

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

Angle of Depression and Angle of Elevation

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

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

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

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

Example \(\PageIndex{1}\)

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

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

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

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

Example \(\PageIndex{2}\)

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

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

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

Example \(\PageIndex{3}\)

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

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

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

Example \(\PageIndex{4}\)

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

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

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

The kite is about 202.7 feet off of the ground.

Example \(\PageIndex{5}\)

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

It might help to draw a picture.

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

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

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

Review (Answers)

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

Additional Resources

Interactive element.

Video: Trigonometry Word Problems Principles - Basic

Activities: Trigonometry Word Problems Discussion Questions

Practice: Trigonometry Word Problems

Real World: Measuring Mountains

TRIGONOMETRY WORD PROBLEMS WITH SOLUTIONS

Problem 1 :

The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60 °. Find the height of the building.

Draw a sketch.

Here, AB represents height of the building, BC represents distance of the building from the point of observation.

In the right triangle ABC, the side which is opposite to the angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and the remaining side is called adjacent side (BC).

Now we need to find the length of the side AB.

tanθ = Opposite side/Adjacent side

tan60° = AB/BC

√3 x 50 = AB

Approximate value of √3 is 1.732

AB = 50 (1.732)

     AB = 86.6 m

So, the height of the building is 86.6 m.

Problem 2 : 

A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60 ° . Find how far the ladder is from the foot of the wall.

Here AB represents height of the wall, BC represents the distance between the wall and the foot of the ladder and AC represents the length of the ladder.

In the right triangle ABC, the side which is opposite to angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

Now, we need to find the distance between foot of the ladder and the wall. That is, we have to find the length of BC.

tanθ = opposite side/adjacent side

BC = (6/√3) x (√3/√3)

BC = (6√3)/3

Approximate value of √3 is 1.732.

BC = 2 (1.732)

BC = 3.464 m 

So,  the distance between foot of the ladder and the wall is 3.464 m.

Problem 3 :

A string of a kite is 100 meters long and t he inclination of the string  with the ground is  60°. Find the height of the kite, assuming that there is no slack in the string.

Here AB represents height of kite from the ground, BC represents the distance of kite from the point of observation.

In the right triangle ABC the side which is opposite to angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

Now we need to find the height of the side AB.

sinθ = opposite side/hypotenuse side

sinθ = AB/AC

sin60° = AB/100

√3/2 = AB/100

(√3/2) x 100 = AB

AB = 50√3 m

So, the height of kite from the ground 50√3 m.

Problem 4 :

From the top of the tower 30 m height a man is observing the base of a tree at an angle of depression measuring 30 ° . Find the distance between the tree and the tower.

Here AB represents height of the tower, BC represents the distance between foot of the tower and the foot of the tree.

Now we need to find the distance between foot of the tower and the foot of the tree (BC).

tan30° = AB/BC

1/√3 = 30/BC

BC = 30(1.732)

 BC = 51.96 m

So, the distance between the tree and the tower is 51.96 m.

Problem 5 :

A man wants to determine the height of a light house. He measured the angle at A and found that tan A = 3/4. What is the height of the light house if A is 40m from the base?

Here BC represents height of the light house, AB represents the distance between the light house from the point of observation.

In the right triangle ABC the side which is opposite to the angle A is known as opposite side (BC), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (AB).

Now we need to find the height of the light house (BC).

tanA = opposite side/adjacent side

tanA = BC/AB

Given : tanA = 3/4.

3/4 = BC/40

Multiply each side by 40.

So, the height of the light house is 30 m.

Problem 6 :

A ladder is leaning against a vertical wall makes an angle of 20° with the ground. The foot of the ladder is 3 m from the wall. Find the length of ladder.

Here AB represents height of the wall, BC represents the distance of the wall from the foot of the ladder.

In the right triangle ABC, the side which is opposite to the angle 20° is known as opposite side (AB),the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

Now we need to find the length of the ladder (AC).

cosθ = adjacent side/hypotenuse side

Cosθ = BC/AC

Cos 20° = 3/AC

0.9397 = 3/AC

AC = 3/0.9396

So, the length of the ladder is about 3.193 m.

Problem 7 :

A kite is flying at a height of 65 m attached to a string. If the inclination of the string with the ground is 31°,  find the length of string.

Here AB represents height of the kite. In the right triangle ABC the side which is opposite to angle 31° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and the remaining side is called adjacent side (BC).

Now we need to find the length of the string AC.

sin31° = AB/AC

0.5150 = 65/AC

AC = 65/0.5150

AC = 126.2 m

Hence, the length of the string is 126.2 m.

Problem 8 :

The length of a string between a kite and a point on the ground is 90 m. If the string makes an angle θ with the ground level such that tan θ = 15/8, how high will the kite be ?

Here AB represents height of the balloon from the ground. In the right triangle ABC the side which is opposite to angle θ is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

tanθ = 15/8 ----> cotθ = 8/15

cscθ = √(1+ cot 2 θ)

cscθ = √(1 + 64/225)

cscθ = √(225 + 64)/225

cscθ = √289/225

cscθ = 17/15 ----> sinθ = 15/17

But, sinθ = opposite side/hypotenuse side = AB/AC.

AB/AC = 15/17

AB/90 = 15/17

So, the height of the tower is 79.41 m.

Problem 9 :

An airplane is observed to be approaching a point that is at a distance of 12 km from the point of observation and makes an angle of elevation of 50°. Find the height of the airplane above the ground.

Here AB represents height of the airplane from the ground. In the right triangle ABC the side which is opposite to angle 50° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

From the figure given above, AB stands for the height of the airplane above the ground.

sin50° = AB/AC

0.7660 = h/12

0.7660 x 12 = h

So, the height of the airplane above the ground is 9.192 km.

Problem 10 :

A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60 °  angle with the ground. Find the  height of the balloon from the ground. (Imagine that there is no slack in the cable)

Here AB represents height of the balloon from the ground. In the right triangle ABC the side which is opposite to angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse (AC) and the remaining side is called as adjacent side (BC).

From the figure given above, AB stands for the height of the balloon above the ground.

sin60° = AB/200

√3/2 = AB/200

AB = (√3/2) x 200

AB = 100(1.732)

AB = 173.2 m

So, the height of the balloon from the ground is 173.2 m.

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Trigonometric word problems, explanations (2).

(Video) Trigonometry Word Problem

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Here's a video by PatrickJMT going over an example of a trigonometric word problem.

Related Lessons

There are a lot of real-world problems that use triangles, so we know we can use trigonometry to solve these types of problems! The most common types of word problems that involve trigonometry are questions involving calculating height, based on other distances/lengths.

Let's walk through an example.

Austin wants to calculate the height of a small tree in his yard. His dad is suggesting that he use a ladder and a tape measure, but Austin thinks he can use trigonometry to find the height of the small tree (which is a lot less work!). Austin tells his dad that if Austin can find the angle of elevation and the length of the tree's shadow, he can use trig to solve the problem. Is Austin right?

Image source: by Anusha Rahman

Austin can measure the length of the base of the triangle, which would be the length of the shadow. He can measure the angle of elevation by drawing an imaginary line from the ground the top tip of the tree, as we can see in the image above.

Now that we've established that Austin can indeed find the height of the tree, let's calculate what that height actually is. At 2pm on a Tuesday, Austin finds that the length of the tree's shadow is 15m. Using a protractor, Austin finds that the angle of elevation of the tree is 30∘.

Let's look back at the image we drew earlier. Can we assign numerical values to the variables?

We have filled in the variables that we know. We have established 15 as the length of the shadow. We also know that the angle of elevation, θ is 30 degrees. Remember to keep track of degrees vs. radians !

Now that we've plugged in the values, we want to solve for the height, which we've established as the variable x. Let's think about what we know - we know the angle and the adjacent side . We want to solve for the opposite side. That sounds like tangent !

tan30∘=x1515⋅tan30∘=x15⋅1√3=x15√3=x

The decimal equivalent of this fraction is 8.66. We have calculated that the height of the small tree in Austin's yard is approximately 8.66 meters tall!

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