Year 11 Extended · Unit 1 Right-angled triangles and circle geometry · Lesson 1 of 7

Two sightings and a bearing: finding a height or a distance you cannot measure

A surveyor cannot reach the base of a minaret, so she sights its top twice. This lesson turns two angles and one walk into a height, then treats a bearing as a pair of components so a journey of several legs can be added up.

Geometry and TrigonometryGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: the legs of a right-angled triangle are 9 cm and 12 cm. Find the hypotenuse.
    Answer15 cm, because 81 + 144 = 225.
  2. From Year 9: a ladder 6 m long leans against a wall at 70° to the ground. How high up the wall does it reach, to 1 decimal place?
    Answer5.6 m, from 6 × sin 70°.
  3. From Year 10: what is the three-figure bearing of due west?
    Answer270°.
Words used in this lesson
angle of elevation
the angle measured up from a level line to a line of sight. For example: looking up at the top of a minaret.
angle of depression
the angle measured down from a level line to a line of sight; it equals the angle of elevation from the ground (alternate angles). For example: looking down from a bridge at a boat.
bearing
a direction measured clockwise from north, written with three figures. For example: 090° is due east.
component
the part of a journey in one direction, such as the east part. For example: walking 5 km east and 3 km north has components 5 and 3.
back bearing
the bearing for the return trip, 180° from the outward bearing. For example: the way home from a landmark.
cotangent
one divided by the tangent of an angle. For example: 1 ÷ tan 40° is the cotangent of 40°.

A tower you cannot walk up to

A surveyor stands at A, level with the base of a minaret, and measures the angle of elevation of its top as 25°. A moat stops her reaching the base. She walks 50 m straight towards the minaret to B and measures again: 40°.

Predict first: is the minaret taller or shorter than the 50 m she walked? Say why, then use the first figure to test your idea. Then change the two sliders and look at what changes the height most.

25
40
52.5 mheight of the tower
62.5 mdistance from B to the foot

Walk 50 m, elevation 25° then 40°: height = 50 ÷ (1/tan 25° − 1/tan 40°) ≈ 52.5 m.

Show me the steps
  1. Walk 50 m, elevation 25° then 40°: the far point is h ÷ tan 25° from the foot and the near point is h ÷ tan 40°. Their difference is 50, so h = 50 ÷ (2.145 − 1.192) ≈ 52.5 m.
  2. Keep 25° and raise the second angle to 70°: the walk of 50 m now covers a longer share of the way in, and the tower is lower, 50 ÷ (2.145 − 0.364) ≈ 28.1 m.
  3. Set the first angle to 35° and the second to 40°: the two sight lines are nearly parallel, so a 50 m walk makes a tiny change of angle and the height jumps to about 211.5 m.
  4. Whatever the two angles, the same relation holds: the walk is the difference of the two cotangents times the height. A small error in either angle is magnified when the angles are close.

Two angles and one walk fix the height. If the foot is a distance x from B, then x + d = h ÷ tan a and x = h ÷ tan b. Subtract: d = h(1/tan a − 1/tan b), so h = d ÷ (1/tan a − 1/tan b).

A journey is a sum of components. A leg of length r on bearing θ has an east part r sin θ and a north part r cos θ, because the bearing is measured clockwise from north. Add the east parts and the north parts separately, then use Pythagoras for the distance and tan−1(east ÷ north) for the angle from north.

60
150
11.7 kmdistance from the port
091°bearing from the port

Legs 10 km on 060° and 6 km on 150°: east 11.7 km, north −0.2 km, so 11.7 km on a bearing of 091°.

Show me the steps
  1. Legs 10 km on 060° and 6 km on 150°: the east parts are 8.66 and 3.00, the north parts 5.00 and −5.20. Totals: east 11.7, north −0.2. The finish is 11.7 km away on 091°, a little south of due east.
  2. Make the second bearing 030°: the east parts are 8.66 and 3.00, the north parts 5.00 and 5.20. Totals: east 11.7, north 10.2, so 15.5 km on 049°.
  3. Make the second bearing 330°: the east part of the second leg is negative, −3.00, so the legs partly cancel in the east direction: east 5.7, north 10.2, and the finish is 11.7 km on 029°.
  4. Make the first bearing 150° and the second 210°: both legs head south, one a little east and one a little west. East 5.0 − 3.0 = 2.0 and north −8.7 − 5.2 = −13.9, so the finish is 14.0 km away on 172°, in the south-east quarter.

The calculator's angle is not always the bearing. tan−1(east ÷ north) is the angle from the north-south line, and it is always between 0° and 90° when the totals are positive. If the north total is negative the point lies south of the port and the bearing is 180° minus that angle (when east is positive); if the east total is negative the bearing is 360° minus the angle, or 180° plus it when the north total is also negative. Sketch before you subtract.

Worked examples

1. From a point on a causeway, the top of a lighthouse is at an angle of elevation of 28°. After walking 60 m straight towards it, the elevation is 52°. How tall is the lighthouse, to 1 decimal place?

  1. Let the lighthouse be h and the second point be x from its foot. Then x = h ÷ tan 52° and x + 60 = h ÷ tan 28°.
  2. Subtract: 60 = h(1 ÷ tan 28° − 1 ÷ tan 52°) = h(1.881 − 0.781) = 1.099h.
  3. h = 60 ÷ 1.099 = 54.6 m.

2. A cargo vessel leaves a harbour and sails 14 km on a bearing of 110°, then 9 km on a bearing of 235°. How far is it from the harbour, and on what bearing, to the nearest degree?

  1. East parts: 14 sin 110° = 13.16 and 9 sin 235° = −7.37, total 5.78. North parts: 14 cos 110° = −4.79 and 9 cos 235° = −5.16, total −9.95.
  2. Distance = √(5.78² + 9.95²) = 11.5 km.
  3. The point is east and south of the harbour. The angle from south is tan−1(5.78 ÷ 9.95) = 30.2°, so the bearing is 180° − 30.2° = 150°.

3. Why does the formula in the key card fail if the second angle equals the first?

  1. If a = b, then 1/tan a − 1/tan b = 0 and h = d ÷ 0 has no value.
  2. Geometrically, of two points on the same side of the tower, the nearer one sees the top at the larger angle, so a walk of d > 0 straight towards the tower always makes b larger than a.
  3. If b is only slightly bigger than a, the denominator is tiny and the height is huge, which is why close angles give unreliable heights.

Your go. The first step is given. From a point on a bridge, the top of a pylon has an angle of elevation of 20°. After walking 45 m straight towards the pylon the elevation is 35°. How tall is the pylon, to 1 decimal place?

  1. The distance from the first point is h ÷ tan 20° and from the second is h ÷ tan 35°, and the two differ by 45.
  2. Write 45 = h(1 ÷ tan 20° − 1 ÷ tan 35°) and find the two cotangents.
Check the answer

45 = h(2.747 − 1.428) = 1.319h, so h = 34.1 m. Treating 45 as the distance from the second point, 45 × tan 35° = 31.5, ignores that the 45 m is only the part of the way between the two points.

With a GDC (optional)

Store the two cotangents before subtracting: type 1 ÷ tan of the first angle, press the store key, then do the same for the second angle. Subtracting two stored numbers avoids the rounding that creeps in when you copy a four-figure value down. Check the display is in degrees (DEG) before any angle is typed.

To turn a pair of totals into a bearing, find the angle from the north-south line with tan−1 of (east ÷ north), then decide from the signs which quarter of the compass the point is in. A scientific calculator does the same work.

Check before you go on

1. From a point A the angle of elevation of the top of a pylon is 21°. From a point B, 35 m closer to it in a straight line, the angle is 38°. What is the height of the pylon, to 1 decimal place?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. An observer whose eyes are 1.6 m above the ground stands 80 m from the foot of a water tower in Dhaka. The angle of elevation of the top of the tower from her eyes is 33°. How tall is the tower, to 1 decimal place?

3. A ferry leaves a port in the Philippines and sails 18 km on a bearing of 062°. How far east of the port is it, to 1 decimal place?

Level 2 · Building

4. From a point A on level ground, the angle of elevation of the top of a cliff is 18°. From a point B, 120 m nearer along the same straight line, it is 31°. How high is the cliff, to the nearest metre?

5. A boat sails 7 km on a bearing of 040°, then 9 km on a bearing of 130°. How far is it from its start, to 1 decimal place?

6. A fishing boat sails 8 km on a bearing of 050°, then 6 km on a bearing of 120°. What is the bearing of its start from where it ends, to the nearest degree?

Level 3 · Stretch

7. Two lookouts A and B stand on a straight shore that runs east to west, 300 m apart, with B to the east of A. A boat is seen from A on a bearing of 040° and from B on a bearing of 320°. How far north of the shore line is the boat, to the nearest metre?

8. From a window of a tower block, two cars in line with its foot are seen at angles of depression of 22° and 41°. The cars are 35 m apart. How far is the nearer car from the foot of the tower block, to 1 decimal place?

Something to think about

Use the key card's formula with a walk of 50 m and angles of 30° and 35°. A surveyor's 35° might really be 36°. Does the height change by a little or by a lot? Then set the first figure to 25° and 70°: why is that height stable?

One way to think about it

By a lot: with 30° and 35° the height is 50 ÷ (1.732 − 1.428) ≈ 165 m, and with 30° and 36° it is 50 ÷ (1.732 − 1.376) ≈ 141 m. When the angles are close, the difference of the cotangents is small, so dividing by it magnifies any error. With 25° and 70° the difference is large and the height is stable.

Exit ticket

9. The bearing of a lighthouse L from a ship S is 312°. What is the bearing of the ship from the lighthouse?