Question

The angles of depression of two ships A and B as observed from the to
light house 60 m high are 60° and 45° respectively. If the two ships are
opposite sides of the light house, find the distance between the two ship
your answer correct to the nearest whole number.​

Answer 1

The distance between the two ships is 20√3(1+√3) m

Answer 2

Let the ship to the left of the light house be named as ship A and to the right of the light house be named as ship B.

Given:

• The height of the light house is 60 m.

• Angle of depression towards ship A is 60°. Angle ACD = 90° - 60° = 30°.

• Angle of depression towards ship B is 45°.

For finding the distance between the ships we need to find the sum of the bases of the 2 right triangles so formed as shown in the diagram.

We need to use the trigonometric ratio tan or cot as they are the ratios which involve the bases of the triangles and the known side.

Using tan:

In triangle ACD:

tanC° = AD/60.

tan 30° = AD/60.

1/√3 = AD/60.

AD = 60/√3 m.

Rationalizing 60/√3:

$\frac{60}{ \sqrt{3} } = \frac{60 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } = \frac{60 \sqrt{3} }{3} = 20 \sqrt{3}$

AD = 20√3 m.

In triangle BCD:

tanC° = BD/60.

tan 45° = BD/60.

1 = BD/60.

BD = 60 m.

Using cot:

In triangle ACD:

cotC° = 60/AD

cot 30° = 60/AD.

√3 = 60/AD.

AD = 60/√3 m.

AD = 20√3 m [After rationalizing].

In triangle BCD:

cotC° = 60/BD.

cot 45° = 60/BD.

1 = 60/BD.

BD = 60 m.

The distance between the ships is AD + BD = AB.

AB = 20√3 + 60 = 20(√3 + 3)m = 20√3(1 + √3) m.

Therefore, the distance between the ships is 20√3(1 + √3) m.

Let's take √3 as 1.73.

Then, 20√3(1+√3):

= 20(1.73)(1+(1.73)).

= 34.6*2.73.

= 94.458.

= 94 m.

Therefore, the distance between the ships is 94 m.