Right-Angled Triangles

Lessons

Senior 3 · Mathematics

Right-Angled Triangles

5 lessons

Lesson 5 of 5

Applications: angles of elevation and depression

25 min

Measuring what you cannot reach

How tall is the tree outside your school? You cannot climb it with a tape measure, but you can stand at a measured distance and measure one angle. Surveyors, foresters and engineers do exactly this, with instruments that read angles above and below the horizontal. This lesson puts the three ratios to work on heights and distances.

Definition

Line of sight: the straight line from an observer's eye to the object being looked at.

Angle of elevation: the angle between the horizontal and the line of sight to an object above the observer.

Angle of depression: the angle between the horizontal and the line of sight to an object below the observer.

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

Example: the angle of depression from P to the boat equals the angle of elevation from the boat to P: they are alternate angles between the parallel horizontal lines.

Important

Solving height and distance problems

Draw a sketch with the horizontal, the line of sight and the right angle.
Mark the angle of elevation or depression from the horizontal.
Find a right-angled triangle containing that angle, one known side and the wanted side.
Choose sine, cosine or tangent, and solve.
Add any extra height, such as the observer's eye level.

Worked example 1: the height of a tree

A pupil stands 25 m from a tree on level ground. Her eye is 1.5 m above the ground, and the angle of elevation of the top of the tree is $32^\circ$ . Find the height of the tree.

The right angle is at K, level with the eye, not at the foot of the tree.

Solution In triangle $E K T$ , $E K = 25$ m is adjacent to $32^\circ$ and $h$ is opposite, so

h = 25\tan 32^\circ \approx 25 \times 0.6249 \approx 15.62\ \text{m}.

Height of the tree $= 15.62 + 1.5 \approx 17.1$ m.

Worked example 2: an angle of depression

From the top of a hill 120 m above a lake, the angle of depression of a boat is $18^\circ$ . How far is the boat from the foot of the hill?

The angle of elevation of P from the boat is also 18°.

Solution In triangle $P Q B$ , the angle at $B$ is $18^\circ$ (alternate angles). The height 120 m is opposite it and $d$ is adjacent:

\tan 18^\circ = \frac{120}{d} \quad\Longrightarrow\quad d = \frac{120}{\tan 18^\circ} \approx 369\ \text{m}.

The direct line-of-sight distance is $PB = \frac{120}{\sin 18^\circ} \approx 388$ m.

Worked example 3: two angles, one height

From a point $A$ on level ground, the angle of elevation of the top of a tower is $30^\circ$ . From $B$ , 20 m closer to the tower, it is $45^\circ$ . Find the height of the tower.

Solution Let $B F = x$ . From $B$ : $\tan 45^\circ = \frac{h}{x}$ , so $h = x$ . From $A$ : $\tan 30^\circ = \frac{h}{x + 20}$ , so $h = \frac{x + 20}{\sqrt{3}}$ .

x = \frac{x + 20}{\sqrt{3}} \;\Longrightarrow\; \sqrt{3}\,x - x = 20 \;\Longrightarrow\; x = \frac{20}{\sqrt{3} - 1} \approx 27.3.

The tower is about 27.3 m high. Neither distance to the foot of the tower was known; two angles and the 20 m between them were enough.

Worked example 4: finding an angle of depression

A security guard at a window 15 m above the ground watches a car parked 40 m from the foot of the building. Find the angle of depression of the car.

Solution The height 15 m is opposite and the 40 m distance adjacent to the angle at the car, which equals the angle of depression: $\tan\theta = \frac{15}{40} = 0.375$ , so $\theta \approx 20.6^\circ$ .

Warning

Measure from the horizontal. The angle of depression is between the horizontal line through the observer and the line of sight, not between the cliff face and the line of sight. In Worked example 2, the angle inside the triangle at $P$ is $90^\circ - 18^\circ = 72^\circ$ .

Don't forget the observer's height when the angle is measured from eye level, as in Worked example 1.

Tip

The sketch is half the answer. Mark the horizontal with a dashed line and put the angle between it and the line of sight. Most exam errors in this topic come from an angle placed against the vertical.

FuseSchool: Angles of Elevation and Depression. Watch where each angle is measured from, and how a right-angled triangle is drawn from the words of a problem.

✦

Practice

Draw a sketch for each problem. Give answers to 3 significant figures or 1 decimal place for angles. The last three are harder.

From a point 40 m from the foot of a building, the angle of elevation of its top is $35^\circ$ . Find the height of the building.
From the top of a cliff 80 m high, the angle of depression of a boat is $25^\circ$ . How far is the boat from the foot of the cliff?
A flagpole 12 m tall casts a shadow 9 m long on level ground. Find the angle of elevation of the sun.
A kite string 60 m long makes an angle of $50^\circ$ with the horizontal, and is held 1.2 m above the ground. How high is the kite?
A boy whose eye is 1.5 m above the ground stands 30 m from a tree and sees its top at an angle of elevation of $28^\circ$ . How tall is the tree?
From the top of a 20 m tower, the angles of depression of two cars on a straight road on the same side of the tower are $45^\circ$ and $30^\circ$ . How far apart are the cars?
Explain why the angle of depression from a lighthouse to a ship equals the angle of elevation of the lighthouse from the ship.

Show answersHide answers

$40\tan 35^\circ \approx 28.0$ m.
$\frac{80}{\tan 25^\circ} \approx 172$ m.
$\tan^{-1}\left(\frac{12}{9}\right) \approx 53.1^\circ$ .
$60\sin 50^\circ + 1.2 \approx 46.0 + 1.2 = 47.2$ m.
$30\tan 28^\circ + 1.5 \approx 15.95 + 1.5 \approx 17.5$ m.
Distances from the foot: $\frac{20}{\tan 45^\circ} = 20$ m and $\frac{20}{\tan 30^\circ} \approx 34.6$ m, so the cars are about 14.6 m apart.
The horizontal through the top of the lighthouse and the sea surface are parallel, and the line of sight crosses both. The two angles are alternate angles, so they are equal.

1Measuring what you cannot reach

Measuring what you cannot reach