(i) The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60⁰ and 45⁰ respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.

(ii) An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45⁰ and 60⁰ respectively. Find the width of the river. Write the answer correct to the nearest whole number.

Solution:

(i) Consider AD as the height of the light house CD = 60 m

Take AD = x m and BD = y m

In triangle ACD

tan 60⁰ = CD/AD

Substituting the values

√3 = 60/x

So we get

x = 60/√3

Multiply and divide by √3

x = 60/√3 × √3/√3 = 60√3/3

x = 20 × 1.732 = 34.64 m

In triangle BCD

tan 45⁰ = CD/BD

Substituting the values

1 = 60/y

y = 60 m

Here the distance between two ships = x + y

= 34.64 + 60

= 94.64 m

= 95 m

(ii) In triangle OMA

tan 45⁰ = OM/AM

Substituting the values

1 = 250/x

So we get

x = 250 m

In triangle OMB

tan 60⁰ = 250/y

Substituting the values

√3 = 250/y

So we get

y = 250/√3 = 250/1.73

y = 144.34

Here

Width of the river = x + y

Substituting the values

= 250 + 144.34

= 394.34 m