Surveyors measure angles up to rooftops, coastguards measure angles down to boats, and navigators steer by bearings. Each one hides a right-angled triangle. Elevation, depression and shortest distance are Extended; bearings are for everyone.
Angles of elevation and depression are measured from the HORIZONTAL, and by alternate angles they are equal. Bearings are measured clockwise from NORTH and written with three figures. Find the right-angled triangle, then use SOH CAH TOA or Pythagoras.
Fig. 1: drawn to scale. The angle of elevation from the observer and the angle of depression from the top are both 32° (alternate angles), and h = 50 tan 32° = 31.2 m
Section 01
Looking up, looking down (Extended).
In the lab, walk the observer towards the tower and watch the angle of elevation grow. Then climb to the top and look down: the angle of depression is measured from your horizontal line of sight, and it always matches the angle of elevation from below.
Angle of elevation: the angle UP from the horizontal to the line of sight.
Angle of depression: the angle DOWN from the horizontal to the line of sight.
They are equal: the two horizontals are parallel, so the angles are alternate angles (Fig. 1).
Put the angle inside the right-angled triangle at the ground end, then use SOH CAH TOA as usual.
Worked example Height from an angle of elevation (Extended)
From a point on level ground 50 m from the foot of a tower, the angle of elevation of the top is \(32^\circ\). Find the height of the tower.
The height is opposite the \(32^\circ\) angle and the 50 m is adjacent: TOA.
Worked example Distance from an angle of depression (Extended)
From the top of a cliff 80 m high, the angle of depression of a boat is \(18^\circ\). How far is the boat from the foot of the cliff?
By alternate angles, the angle of elevation of the clifftop from the boat is also \(18^\circ\).
In the triangle at the boat: the height 80 m is opposite \(18^\circ\) and the distance \(d\) is adjacent, so \(\tan 18^\circ = \dfrac{80}{d}\).
\(d = \dfrac{80}{\tan 18^\circ} = 246.21\ldots\)
Answer: \(246\) m (3 s.f.)
Your turn
(Extended) From a point 30 m from the base of a tree, the angle of elevation of the top is \(35^\circ\). Find the height of the tree.
\(h = 30\tan 35^\circ = 21.0\) m (3 s.f.).
Examiner trap
Measuring the angle of depression from the vertical cliff face. It is always measured from the horizontal. If you put \(18^\circ\) at the top of the cliff against the vertical, you get \(80\tan 18^\circ = 26.0\) m, which is far too close.
Section 02
Bearings with trigonometry.
A bearing is an angle measured clockwise from north, written with three figures (\(060^\circ\), not \(60^\circ\)). Draw a north line at every point you work from. Splitting a journey into its north-south and east-west parts makes a right-angled triangle.
A ship sails 25 km on a bearing of \(060^\circ\). How far north and how far east of its starting point is it?
The \(60^\circ\) is between the north line and the path, so north is adjacent and east is opposite. The 25 km path is the hypotenuse.
North: \(25\cos 60^\circ = 12.5\) km.
East: \(25\sin 60^\circ = 21.65\ldots\) km.
Answer: \(12.5\) km north and \(21.7\) km east (3 s.f.)
Worked example Finding a bearing
Town \(B\) is 8 km east and 5 km south of town \(A\). Find the bearing of \(B\) from \(A\), and the bearing of \(A\) from \(B\).
At \(A\), the angle below the east line is \(\tan^{-1}\left(\dfrac{5}{8}\right) = 32.005\ldots^\circ\).
East is \(090^\circ\), so the bearing of \(B\) from \(A\) is \(90^\circ + 32.0^\circ = 122.0^\circ\).
Back bearing: \(122.0^\circ + 180^\circ = 302.0^\circ\).
Answer: \(122.0^\circ\) and \(302.0^\circ\)
Worked example A journey with a right-angle turn
A boat sails 30 km on a bearing of \(040^\circ\), then 20 km on a bearing of \(130^\circ\). Find its distance and bearing from the start.
\(130^\circ - 40^\circ = 90^\circ\), so the two legs meet at a right angle.
Distance: \(\sqrt{30^2 + 20^2} = \sqrt{1300} = 36.055\ldots\) km.
Angle between the first leg and the line back to the start: \(\tan^{-1}\left(\dfrac{20}{30}\right) = 33.69\ldots^\circ\).
Bearing: \(40^\circ + 33.7^\circ = 73.7^\circ\).
Answer: \(36.1\) km on a bearing of \(073.7^\circ\)
Examiner trap
Measuring anticlockwise, or from the wrong point. 'The bearing of \(B\) from \(A\)' means stand at \(A\), face north, turn clockwise until you face \(B\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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What is the three-figure bearing of south-west?
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The bearing of \(B\) from \(A\) is \(070^\circ\). What is the bearing of \(A\) from \(B\)?
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(Extended) From a point 20 m from the foot of a tree, the angle of elevation of the top is \(25^\circ\). How tall is the tree?
Quick check+10 XP
(Extended) Why is the angle of depression from a clifftop to a boat equal to the angle of elevation from the boat to the clifftop?
Section 03
The shortest distance (Extended).
The shortest distance from a point to a straight line is along the perpendicular. This is how far a lighthouse is from a ship's path at the closest moment.
Worked example Closest approach (Extended)
A lighthouse \(L\) is 12 km due east of a harbour \(H\). A ship leaves \(H\) on a bearing of \(070^\circ\). Find the shortest distance between the lighthouse and the ship's path.
\(HL\) is on a bearing of \(090^\circ\), so the angle between \(HL\) and the path is \(90^\circ - 70^\circ = 20^\circ\).
Drop a perpendicular from \(L\) to the path, meeting it at \(P\). Triangle \(HPL\) has hypotenuse \(HL = 12\) km, and \(LP\) is opposite the \(20^\circ\).
\(LP = 12\sin 20^\circ = 4.104\ldots\) km
Answer: \(4.10\) km (3 s.f.)
Examiner trap
Using the distance along the path, \(12\cos 20^\circ = 11.3\) km, or the original 12 km. The shortest distance is the perpendicular one, and it is always less than the distance to \(H\).
Stage 2 · Exam
Exam-style questions
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(Extended) From the top of a cliff 45 m high, the angle of depression of a boat is \(14^\circ\). Find the distance of the boat from the foot of the cliff.
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A ship sails 18 km from port \(P\) on a bearing of \(125^\circ\). How far east of \(P\) is the ship?
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\(B\) is 6 km west and 9 km north of \(A\). Find the bearing of \(B\) from \(A\).
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(Extended) A lighthouse \(L\) is 15 km due north of a harbour \(H\). A ship leaves \(H\) on a bearing of \(050^\circ\). Find the shortest distance from \(L\) to the ship's path.
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