This note covers angles of elevation and depression, three-figure bearings, and how to combine them with right-angled and non-right-angled triangle methods to solve real distance and height problems. These skills are used in surveying, aviation, sailing and engineering.
Key ideas
- The angle of elevation is the angle measured upwards from the horizontal to a line of sight. The angle of depression is the angle measured downwards from the horizontal.
- The angle of depression from point \(A\) down to point \(B\) equals the angle of elevation from \(B\) up to \(A\). The two horizontal lines are parallel, so these are alternate angles.
- A bearing is an angle measured clockwise from north, written with three figures. For example, east is \(090^\circ\), south is \(180^\circ\), and \(045^\circ\) is north-east.
- To find a back bearing (the bearing of \(A\) from \(B\) when you know the bearing of \(B\) from \(A\)), add \(180^\circ\) if the first bearing is less than \(180^\circ\), and subtract \(180^\circ\) if it is more.
- Always draw a north line at each point where a bearing is measured. North lines are parallel, so angles between a path and the north line can be matched using co-interior or alternate angles.
- Real problems often need several steps: a right-angled triangle for a height, then the sine or cosine rule for a distance in a non-right-angled triangle.
A good method is: sketch a diagram with all given information, mark the angle you need, find a triangle that contains it, then choose the right rule.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Angle of elevation | Angle above the horizontal when looking up at an object. |
| Angle of depression | Angle below the horizontal when looking down at an object. |
| Three-figure bearing | Clockwise angle from north, from \(000^\circ\) to \(360^\circ\). Write leading zeros, such as \(075^\circ\). |
| Back bearing | The reverse direction: bearing \(\pm 180^\circ\). |
| \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\) | Height and horizontal distance problems. |
| Sine rule, cosine rule, \(\frac{1}{2}ab\sin C\) | Triangles with no right angle, such as ship and flight paths. |
Worked examples
Example 1 A surveyor stands 60 m from the base of a communications tower on level ground. Her eye is 1.7 m above the ground and the angle of elevation of the top of the tower is \(35^\circ\). Find the height of the tower.
The height of the tower above eye level is opposite the \(35^\circ\) angle, and 60 m is adjacent.
\(\tan 35^\circ = \dfrac{h}{60}\), so \(h = 60 \tan 35^\circ \approx 42.0\) m.
Add the height of her eye: \(42.0 + 1.7 = 43.7\) m. (Using the unrounded value gives \(43.71\) m.)
Answer: the tower is about 43.7 m tall.
Example 2 A ship sails 12 km on a bearing of \(060^\circ\), then 9 km on a bearing of \(150^\circ\). Find how far the ship is from its starting point and the bearing on which it must sail to return directly.
Draw north lines at the start and at the turning point. The second leg turns clockwise by \(150^\circ - 60^\circ = 90^\circ\), so the two legs meet at a right angle.
Distance: \(d = \sqrt{12^2 + 9^2} = \sqrt{225} = 15\) km.
The angle at the start between the first leg and the line back to the start is \(\theta\), where \(\tan\theta = \dfrac{9}{12}\), so \(\theta = \tan^{-1}(0.75) \approx 36.9^\circ\).
The bearing of the final position from the start is \(60^\circ + 36.9^\circ = 96.9^\circ\).
Back bearing: \(96.9^\circ + 180^\circ = 276.9^\circ\).
Answer: 15 km from the start; return bearing \(277^\circ\) (to the nearest degree).
Example 3 Two observers stand in a straight line with a building. The first is further away and sees the top of the building at an angle of elevation of \(25^\circ\). The second is 50 m closer and sees it at \(40^\circ\). Find the height of the building. Ignore the height of the observers.
Let the height be \(h\). The horizontal distance from the nearer observer is \(\dfrac{h}{\tan 40^\circ}\) and from the further observer it is \(\dfrac{h}{\tan 25^\circ}\).
The difference is 50 m: \(\dfrac{h}{\tan 25^\circ} - \dfrac{h}{\tan 40^\circ} = 50\).
\(h(2.1445 - 1.1918) = 50\), so \(h = \dfrac{50}{0.9527} \approx 52.5\) m.
Check: the nearer distance is \(52.5 \div \tan 40^\circ \approx 62.6\) m, and the further one is \(112.6\) m, giving \(112.6 \tan 25^\circ \approx 52.5\) m. The values agree.
Answer: the building is about 52.5 m tall.
Common errors
- Measuring an angle of depression from the vertical, or from the line of sight itself, instead of from the horizontal.
- Writing a bearing without three figures, such as \(75^\circ\) instead of \(075^\circ\), or measuring it anticlockwise.
- Not drawing a north line at the second point, then using the angle from the first point by mistake.
- Using Pythagoras or right-angled trigonometry in a triangle that has no right angle. Check with the sketch.
- Forgetting to add the height of the observer (or subtract it) when the question gives it.
Exam tip: Mark every north line and every horizontal line on your sketch, and write in each angle you can work out from parallel lines before choosing a formula.
Check yourself
- A 45 m building stands on level ground. Find the angle of elevation of its top from a point on the ground 80 m from its base.
Answer
\(\tan\theta = \frac{45}{80} = 0.5625\), so \(\theta \approx 29.4^\circ\). - From the top of a 30 m balcony, the angle of depression of a parked car on level ground is \(18^\circ\). How far is the car from the foot of the balcony?
Answer
The angle at the car equals \(18^\circ\) (alternate angles). \(\tan 18^\circ = \frac{30}{d}\), so \(d = \frac{30}{\tan 18^\circ} \approx 92.3\) m. - The bearing of a lighthouse from a boat is \(072^\circ\). Find the bearing of the boat from the lighthouse.
Answer
\(072^\circ + 180^\circ = 252^\circ\). - A plane flies 200 km due east, then 150 km due south. Find its final bearing from its start.
Answer
The angle south of east is \(\tan^{-1}\left(\frac{150}{200}\right) \approx 36.9^\circ\). Bearing \(= 090^\circ + 36.9^\circ = 126.9^\circ\), or \(127^\circ\).