Math Core

Surveyors, pilots, sailors and engineers regularly need distances they can't measure with a tape: the height of a cliff, the distance to a ship, the altitude of a plane. What they can measure is an angle. Pair that angle with one known distance and a right triangle does the rest.

Two angles measured from the horizontal

When you look at something, the straight path from your eye to the object is your line of sight. The angle it makes with a horizontal line tells you how steeply you are looking up or down.

Definition

Angle of elevation and angle of depression

The angle of elevation is the angle between the horizontal and your line of sight when you look up at an object.

The angle of depression is the angle between the horizontal and your line of sight when you look down at an object.

Both are measured from the horizontal, never from the vertical.

The observer on the cliff looks down at the boat (angle of depression). Someone in the boat looks up at the cliff top (angle of elevation). The two angles are equal.

Suppose a person at the top of a cliff looks down at a boat, and someone in the boat looks up at the cliff top. They are looking along the same line of sight. The horizontal line through the cliff top (dashed) is parallel to the water, and the line of sight is a transversal cutting both. So the angle of depression from the cliff and the angle of elevation from the boat are alternate interior angles, and they are equal.

Depression equals elevation

The angle of depression from $A$ down to $B$ equals the angle of elevation from $B$ up to $A$ . So when you draw the right triangle, you can put the angle of depression at the object on the ground, as an angle inside the triangle.

A plan for every problem

Sketch the situation. Draw the horizontal, the vertical (a building, a tree, an altitude), and the line of sight.
Find the right triangle and mark the angle and the known side.
Choose a ratio with SOH-CAH-TOA, write the equation, and solve.
Check that the answer makes sense and has units.

Most of these problems involve a horizontal distance and a height, which are the two legs. That makes tangent the ratio you'll use most often.

Worked example: Height of a tree

You stand $40$ m from the base of a tree on level ground. The angle of elevation to the top of the tree is $32^\circ$ . How tall is the tree, to the nearest tenth of a meter? (Treat your eye as being at ground level.)

The line of sight from the ground to the treetop rises at 32°.

The height $h$ is opposite the $32^\circ$ angle and the $40$ m distance is adjacent:

\tan 32^\circ = \frac{h}{40} \quad\Longrightarrow\quad h = 40 \tan 32^\circ \approx 40(0.6249) \approx 25.0 \text{ m}.

Worked example: A boat seen from a lighthouse

From the top of a lighthouse $55$ m above the sea, a keeper sees a boat at an angle of depression of $12^\circ$ . How far is the boat from the base of the lighthouse, to the nearest tenth of a meter?

The angle of depression is measured down from the horizontal (dashed) line at the top of the lighthouse.

The angle of elevation from the boat is also $12^\circ$ . In the triangle, the $12^\circ$ angle sits at the boat, the $55$ m height is opposite it and the distance $d$ is adjacent:

\tan 12^\circ = \frac{55}{d} \quad\Longrightarrow\quad d = \frac{55}{\tan 12^\circ} \approx \frac{55}{0.2126} \approx 258.8 \text{ m}.

A small angle of depression means the boat is far away compared with the height, so a distance of almost five times the height is reasonable.

Common mistake

The most common mistake is putting the angle of depression at the top of the triangle, between the line of sight and the vertical tower. That angle is actually $90^\circ - 12^\circ = 78^\circ$ . The angle of depression is measured from the horizontal, so either use $78^\circ$ at the top, or (easier) move the $12^\circ$ down to the object on the ground.

Finding the angle

If you know the two distances, use an inverse trig function to find the angle.

Worked example: Angle of the sun

A $6$ m flagpole casts a shadow $9$ m long on level ground. What is the angle of elevation of the sun, to the nearest tenth of a degree?

The sun's rays run from the top of the pole to the tip of the shadow. At the tip, the pole ( $6$ m) is opposite and the shadow ( $9$ m) is adjacent:

\tan\theta = \frac{6}{9} \quad\Longrightarrow\quad \theta = \tan^{-1}\!\left(\frac{2}{3}\right) \approx 33.7^\circ.

Two angles, one unknown distance

Sometimes you can't reach the base of the object, so you take two sightings from different spots. Each one gives a right triangle, and the two triangles share the same height.

Worked example: Two sightings of a building

From point $Q$ , the angle of elevation to the top of a building is $50^\circ$ . You walk $30$ m directly away from the building to point $P$ , and the angle of elevation drops to $35^\circ$ . How tall is the building, to the nearest tenth of a meter?

From Q the top of the building is at 50°. From P, 30 m farther back, it is at 35°.

Let $h$ be the height and $x$ the distance from $Q$ to the building. The two right triangles give

h = x \tan 50^\circ \qquad\text{and}\qquad h = (x + 30)\tan 35^\circ.

Set them equal and solve for $x$ :

\begin{aligned} x \tan 50^\circ &= x \tan 35^\circ + 30 \tan 35^\circ \\ x(\tan 50^\circ - \tan 35^\circ) &= 30 \tan 35^\circ \\ x &= \frac{30 \tan 35^\circ}{\tan 50^\circ - \tan 35^\circ} \approx \frac{21.006}{0.4915} \approx 42.74 \end{aligned}

Then $h = x \tan 50^\circ \approx 42.74(1.1918) \approx 50.9$ m.

Tip

When an observer's eyes are above the ground, the triangle starts at eye level, not at the ground. Find the height above your eyes with trig, then add your eye height at the end.

Practice

Round lengths to the nearest tenth and angles to the nearest tenth of a degree unless told otherwise.

Practice 1

A kite string is $80$ ft long and makes an angle of elevation of $55^\circ$ with the ground. Assuming the string is straight, how high is the kite above the point where the string is held (treat that point as ground level)?