Trigonometry Worksheets

Practice and master trigonometry concepts with our helpful walkthroughs and downloadable practice worksheets from our team of elite math educators.

Angle of elevation and depression

By Vighnesh Hemnani

Angles of elevation and depression are extremely integral in order to understand how and where certain objects are positioned relative to others. Here, we discover what these angles are, how to find them, how they are being used via solved examples and practice problems.

Why is this concept useful?

This concept is very useful in real life because it can help find distances between objects and/or speeds of objects. Some examples include control/radar stations mapping the nearby planes, lighthouses watching over boats approaching the shore or someone just looking up at a cat in a tree.

Where does this concept fit into the curriculum?

High School Geometry

What is an angle of elevation/depression?

Let’s say that you are looking at a plane flying in the sky. In this example, the angle of elevation would be the angle made between the object and the horizontal line made at your eye-level, like in the diagram below:

On the other hand, an angle of depression can be considered its reverse: if you were the pilot looking down at a person on land, then the angle of depression is the angle between the person on the land and the horizontal line at your eye-level, like in the diagram above.

How to use this concept?

There are many ways one can use this concept. To begin, when an angle of elevation or depression is made, a right-angled triangle is also made, making concepts of trigonometry and Pythagorean theorem applicable. However, in angle of elevation and depression problems, trigonometry is more of a used concept since it includes angles.

If we are given the angle of elevation or depression in a problem and one of the sides of the triangle, we can use trigonometry to find other sides or lengths of the given triangle. Sometimes, trigonometry and angle of elevation/depression can be an intermediate step to find other distances such as the distance of a bird in a tree or the distance between boats.

Equations pertaining the red and blue triangles both stem from trigonometry except the difference is that each triangle reference different sides and the angles are different,

θ\theta
and
α\alpha
. However, the trig. equations hold in the same way. Below are a set of equations for the blue triangle with angle
θ\theta
:

sin (θ) =opphyp      cos (θ) =adjhyp        tan (θ)=oppadj sin\ \left(\theta\right)\ =\frac{opp}{hyp}\ \ \ \ \ \ cos\ \left(\theta\right)\ =\frac{adj}{hyp}\ \ \ \ \ \ \ \ tan\ \left(\theta\right)=\frac{opp}{adj}\

So, as you can see, knowing two of the variables, you can easily solve for another variable. Sometimes, you might be asked to find the angle of elevation/depression or a particular distance between two objects.

Sample Math Problems

Problem

Elon Musk just launched a rocket into the sky, and he is watching the take off several kilometers away from the launchpad. Given that the rocket has travelled 1 km from the launchpad (which is at Elon’s eye-level) and he knows the angle of elevation is 20°, how far away is he from the launch pad?

Solution

2.75 km. From the question, we can draw the following right-angled triangle to map the distances:

Based on the above image, we can use trigonometry to find the missing side, specifically the tangent relation:

tan (θ) =oppadjtan\ \left(\theta\right)\ =\frac{opp}{adj}
tan (20)=1kmxtan\ (20)= \frac{1km}{x}
x=1kmtan (20)=2.7475…km≈2.75kmx = \frac{1km}{tan\ (20)} = 2.7475…km ≈ 2.75 km

Problem

The plane, Air Force One, is at an altitude of 1250 miles cruising over New York City ready to land. It sees the control tower at JFK at an angle of depression of 34°. What is the distance between Air Force One and the control tower?

Solution

2,235 miles. Based on the question, we can draw the following right-angled triangle to map the distances:

Now, using the triangle, we know we need to solve for the distance between the plane and the tower, so we can use trigonometry, specifically the sine relation:

sin (θ) =opphypsin\ \left(\theta\right)\ =\frac{opp}{hyp}
sin 34=1250mixsin\ 34 = \frac{1250mi}{x}
x=1250misin(34)=2,235.3645mi≈2,235mix=\frac{1250mi}{sin(34)} = 2,235.3645 mi ≈ 2,235 mi

Problem

A 4.5 m ladder is kept leaning against a 3.6 m tall wall. What should the angle of elevation from the foot of the ladder be in order for the ladder to be able to reach the top of the wall?

Solution

53°.

The first step to solving most geometry questions entails in drawing a reasonably precise sketch of the situation to understand the missing piece better. In this case, the ladder, the wall and the ground make a right-angled triangle like so:

Knowing this, we can apply the sine relation of trigonometry to find the angle of elevation:

sin(θ) =opphypsin\left(\theta\right)\ =\frac{opp}{hyp}
sin θ=3.6m4.5msin\ θ = \frac{3.6m}{4.5m}
θ=(3.6m4.5m) =53.13≈53degrees\theta=\left(\frac{3.6m}{4.5m}\right)\ =53.13≈53 degrees

Problem

LeBron, who is 6 feet and 9 inches, flies a kite and lets out 88 feet of string. Given that it was a windy day and the string has stretched out fully and the angle of elevation is 61°, how high is the kite off the ground?

Solution

83.75 ft. Again, the first step is to sketch the situation, which has been done as below:

Now, if we are to find out how high the kite off the ground is, we need to find out how much the kite has flown vertically from LeBron’s eye-level. This means finding the opposite side to the angle of elevation. To do this, we will use the sine relation of trigonometry:

sin(θ) =opphypsin\left(\theta\right)\ =\frac{opp}{hyp}
sin(61)=opp88ftsin(61) = \frac{opp}{88ft}
opp=sin(61)×88ft=76.965≈77ftopp=sin(61) × 88ft = 76.965 ≈ 77ft

Looking back at the question, they are asking for the height of the kite from the ground – this would imply we need to add LeBron’s height to 77 ft. Note that 6 feet 9 inches doesn’t translate to 6.9 ft but translates to 6.75 ft (9 in of a foot/out of 12 in is 0.75). So, the final answer is: 77 + 6.75 = 83.75 ft.

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Practice Math Problems

1. With a few friends, you are visiting a cliff on the shore. From Google, you know that the cliff is located 450 m above sea level. Then, you and your friends spot two emergency lifeboats floating in the distance. Using your tools, you know that the angles of depression for the two lifeboats are 4° and 12°. How far are the two lifeboats from each other?

2. With a few friends, you are visiting a cliff on the shore. From Google, you know that the cliff is located 450 m above sea level. Then, you and your friends spot an emergency lifeboat floating in the distance. Using your tools, you know that the angles of depression for the lifeboats are 7°. How far is the lifeboat from the shore?

3. 17m from the base of a flagpole, it is noted that the angle of elevation of the top of the pole is 48°. Then, how tall is the flagpole?

4. Your 6 ft brother, Sherlock, is flying a small remote-controlled airplane in a park. Say you know that the altitude of the RC airplane is 70 ft, and the angle of elevation from your brother’s eye-level is 50°. Then, how far away is the airplane from your brother?

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