Question

As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships

Answer 1

Answer:

As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use √3 = 1.732]

Step-by-step explanation:

Let AB be the height of the light house and two ships be at C and D.

In ∆ABC,

tan 45° = AB/BC

1 = 100 / BC

BC = 100 m

In ∆ABD,

tan 30° = AB/BD

1/√3 = 100 / (BC + CD)

1/√3 = 100 / (100 + CD)

100 + CD = 100√3

CD = 100√3 - 100

CD = 100 × (√3 - 1)

Given √3 = 1.732

CD = 100 × (1.732 - 1)

CD = 100 × 0.732

CD = 73.2 m

Hence, the distance between the two ships is 73.2 m

Answer 2

Answer:

the distance between the two ships is 73.2