IGCSE · Practice questions · Elevation, depression and bearings

North, clockwise, three figures.

Six original Cambridge-style questions on angles of elevation and depression (Extended), bearings, and the shortest distance to a line (Extended).

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Sketch it, find the right angle, then trigonometry.

Sketch: draw a clear diagram with the horizontal ground or north lines marked.
Elevation and depression: both measured from the horizontal, and equal by alternate angles.
Bearings: clockwise from north, three figures; back bearing is \(\pm 180^\circ\).
Journeys: split into north-south and east-west parts, or spot a right-angle turn.
Shortest distance: the perpendicular from the point to the line.

Recall

[3 marks]

(a) Write down the three-figure bearing of due west.

(b) Write down the three-figure bearing of south-east.

(a) \(270^\circ\) ✓
(b) \(135^\circ\) ✓
(c) \(312^\circ - 180^\circ = 132^\circ\) ✓

Calculation

[2 marks]

(Extended) A building is 24 m tall. From a point \(P\) on level ground, the angle of elevation of the top of the building is \(37^\circ\). Calculate the distance from \(P\) to the foot of the building.

\(\tan 37^\circ = \dfrac{24}{d}\) ✓
\(d = \dfrac{24}{\tan 37^\circ} = 31.8\) m (3 s.f.) ✓

Calculation

[2 marks]

A ship sails 40 km from port on a bearing of \(150^\circ\). Calculate how far (a) south and (b) east of the port the ship is.

The angle between the path and the south line is \(180^\circ - 150^\circ = 30^\circ\). ✓
(a) South: \(40\cos 30^\circ = 34.6\) km (3 s.f.) ✓
(b) East: \(40\sin 30^\circ = 20\) km ✓

Calculation

[4 marks]

A plane flies from \(A\) 120 km due east to \(B\), then 90 km due north to \(C\).

(a) Calculate the distance \(AC\).

(b) Find the bearing of \(C\) from \(A\).

(a) \(AC = \sqrt{120^2 + 90^2} = \sqrt{22\,500} = 150\) km ✓
(b) Angle north of east at \(A\): \(\tan^{-1}\left(\dfrac{90}{120}\right) = 36.87^\circ\). Bearing \(= 90^\circ - 36.87^\circ = 053.1^\circ\). ✓
(c) \(53.1^\circ + 180^\circ = 233.1^\circ\) ✓

Analysis

[3 marks]

(Extended) From the top of a lighthouse 35 m above sea level, the angles of depression of two boats are \(12^\circ\) and \(20^\circ\). The boats and the foot of the lighthouse are in a straight line, with both boats on the same side.

Calculate the distance between the two boats.

Far boat (\(12^\circ\)): distance \(= \dfrac{35}{\tan 12^\circ} = 164.66\ldots\) m ✓
Near boat (\(20^\circ\)): distance \(= \dfrac{35}{\tan 20^\circ} = 96.16\ldots\) m ✓
Distance between boats \(= 164.66\ldots - 96.16\ldots = 68.5\) m (3 s.f.) ✓

Analysis

[4 marks]

(Extended) A rock \(R\) is 10 km due east of a harbour \(H\). A boat leaves \(H\) and sails in a straight line on a bearing of \(058^\circ\).

(a) Calculate the shortest distance between the rock and the boat's path.

(b) How far has the boat sailed when it is closest to the rock?

Angle between \(HR\) (bearing \(090^\circ\)) and the path: \(90^\circ - 58^\circ = 32^\circ\). ✓
(a) Perpendicular distance \(= 10\sin 32^\circ = 5.30\) km (3 s.f.) ✓
(b) Distance along the path \(= 10\cos 32^\circ = 8.48\) km (3 s.f.) ✓
(c) \(5.30 > 5\), so yes, the boat stays more than 5 km from the rock. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.