10 min readPathway sheet3-mark question, or a 5-mark long answer

Two right triangles sharing one height

When two angles are seen along one straight line, both triangles share the same height h. Write tan for each triangle and solve the two equations together.

Shared sideThe vertical height h appears in both triangles

Same side of the towerDistance between the points = larger distance − smaller distance

Opposite sides of the towerDistance between the points = sum of the two distances

Shortcut for 30° and 60°, same side, d aparth = (√3/2)d, and the nearer point is d/2 from the foot

How to draw the figure

Draw the tower as a vertical line AB with foot B. Mark the right angle at B.
Same side: put points C and D on the ground on one side of B, with D farther away. Join AC and AD.
Mark the larger angle at the nearer point C and the smaller angle at D.
Opposite sides: put C on the left of B and D on the right of B.
Label AB = h and BC = x, and write the given distance CD on the figure.

Method

1
Draw the figure and mark both right triangles.
2
In the nearer triangle, write tan of its angle = h/x. Number it (1).
3
In the other triangle, write tan of its angle = h/(x + d) for the same side, or h/y for the opposite side. Number it (2).
4★
Write h in terms of x from (1) and substitute it in (2).
5
Solve for x, then find h.
6
Use √3 = 1.732 and write units.

★ key step

Same side: walking towards a tower

Problem

The angle of elevation of the top of a tower changes from 30° to 60° as a student walks 40 m towards it. Find the height of the tower. (Use √3 = 1.732)

Working

Let AB = h be the tower. The student starts at D and stops at C, so DC = 40 m. Let BC = x.
In ΔABC: tan 60° = h/x, so h = √3 x ... (1)
In ΔABD: tan 30° = h/(x + 40), so √3 h = x + 40 ... (2)
Put (1) in (2): √3 × √3 x = x + 40, so 3x = x + 40.
2x = 40, so x = 20 m.
h = 20√3 = 20 × 1.732 = 34.64 m.

Answer

The tower is 34.64 m high (the student is now 20 m from its foot).

Opposite sides: two observers

Problem

Two men stand on opposite sides of a 60 m tall tower, in line with its foot. They see its top at angles of elevation of 30° and 60°. Find the distance between them. (Use √3 = 1.732)

Working

Let AB = 60 m be the tower, C and D the two men. ∠ACB = 30°, ∠ADB = 60°.
In ΔABC: tan 30° = AB/BC, so 1/√3 = 60/BC and BC = 60√3 m.
In ΔABD: tan 60° = AB/BD, so √3 = 60/BD and BD = 60/√3 = 20√3 m.
CD = BC + BD = 60√3 + 20√3 = 80√3 m.
CD = 80 × 1.732 = 138.56 m.

Answer

138.56 m

Same side vs opposite sides

Both points on the same sideThe foot of the tower is outside the two points. Distance between them = BD − BC, so write the farther distance as x + d.

Points on opposite sidesThe foot of the tower lies between the two points. Distance between them = BC + BD, so add the two distances.

Read the question for words like 'on the same side' or 'on either side'. They decide whether you add or subtract.

In real life

On a trek in the Sahyadri hills, you can estimate a cliff's height from two angle readings taken 20 m apart. Then you solve two triangles, just like this.

Exam angle

2 marks: “Two points lie on opposite sides of a 30 m pole, in line with its foot. From them, the angles of elevation of its top are 45° and 60°. Find the distance between the points. (Use √3 = 1.732)” Distances: 30/tan 45° = 30 m and 30/tan 60° = 10√3 = 17.32 m. Sum = 47.32 m.

3 marks: “From a point P on the ground, the angle of elevation of the top of a tower is 30°. After walking 20 m towards the tower, it becomes 60°. Find the height of the tower and its distance from P. (Use √3 = 1.732)” h = √3 x and √3 h = x + 20 give 3x = x + 20, so x = 10 m. h = 10√3 = 17.32 m. Distance from P = 30 m.

5 marks: “Two ships are on the same side of a 100 m tall lighthouse, in line with its foot. From the top, the angles of depression of the ships are 30° and 45°. Find the distance between the ships. (Use √3 = 1.732)” Draw the figure with alternate angles. Nearer ship: 100/tan 45° = 100 m. Farther ship: 100/tan 30° = 100√3 = 173.2 m. Distance = 173.2 − 100 = 73.2 m.

Remember it

Same side: subtract. Opposite sides: add. The height h is the same in both triangles.