Coaching Centre
Problems with right-angled triangles
Solve practical problems using angles of elevation and depression, compass bearings and true bearings, drawing a clear diagram first.
Elevation and depression
The angle of elevation is measured up from the horizontal to an object. The angle of depression is measured down from the horizontal. Both are always measured from the horizontal, never from the vertical.
From a point $40$ m from the base of a tower, the angle of elevation of the top is $28^\circ$. Find the height of the tower to the nearest metre.
Bearings
| Type | How it is measured | Example |
|---|---|---|
| True bearing | Clockwise from north, as three digits | $065^\circ\text{T}$, $230^\circ\text{T}$ |
| Compass bearing | From north or south, turning towards east or west | N$65^\circ$E, S$50^\circ$W |
North is up the y-axis. Drag P and read its true bearing from the origin O, measured clockwise from north.
- Put $P$ due east of $O$. What is its bearing?
- Find a point with bearing $225^\circ\text{T}$. What compass bearing is that?
A ship sails $12$ km on a bearing of $070^\circ\text{T}$. How far east and how far north of its start is it?
From $30$ m away, the angle of elevation of a tree top is $35^\circ$. Find its height to the nearest metre.
$30\tan 35^\circ \approx 21.0$ m.
The bearing $\text{S}40^\circ\text{E}$ as a true bearing is:
From south ($180^\circ$), turn $40^\circ$ back towards east: $180 - 40 = 140$.
The bearing of $B$ from $A$ is $060^\circ\text{T}$. The bearing of $A$ from $B$ is:
Add $180^\circ$: $240^\circ\text{T}$.
$300^\circ\text{T}$ as a compass bearing is:
$360 - 300 = 60^\circ$ west of north.
From a cliff $50$ m high, the angle of depression of a boat is $28^\circ$. How far is the boat from the base of the cliff, to the nearest metre?
The angle of elevation of the cliff top from the boat is also $28^\circ$, so $\tan 28^\circ = \dfrac{50}{d}$ and $d = \dfrac{50}{\tan 28^\circ} \approx 94$ m.
A hiker walks $3$ km north, then $4$ km east. How far is the hiker from the start?
$\sqrt{3^2 + 4^2} = 5$ km.
The angle of elevation is measured from:
Elevation and depression are both measured from the horizontal.
A $10$ m ladder reaches $8$ m up a wall. Find the angle it makes with the ground to the nearest degree.
$\sin\theta = \dfrac{8}{10}$, so $\theta \approx 53^\circ$.