Question

As observed from the top of a 60m high light - house from sea-level the angles of depression of two ships are 30and 45. If one ship is exactly behind the other on the same side of the light- house, find the distance between the two ships. [use3=1.732]

Answer 1

Hope it will be helpful to you

Answer 2

Given:

Height of the light = 60m

Angle of depression = 30

Angle of depression = 45

To Find:

The distance between the two ships.

Solution:

Let us take a triangle as ABD, where AC is the diagonal.

Let the two ships be C and D at a distance CD from each other.

The angle of depression of one ship is 45°.

So, ∠ACB = 45°

Similarly, The angle of depression of another ship is 30°.

∠ADB = 30°

Lines AE and BD are parallel, and AD is a transversal,

Thus,

∠ADB = ∠EAD = 30°

Similarly,

∠ACB = ∠EAC = 45°

Since, the lighthouse is perpendicular to the ground,  = ∠ABD = 90°

Now, In ΔABC

Tan45 = AB/BC

1 = 60/BC

BC = 60

In ΔABD

Tan30 = AB/BD

1/√3 = 60/BD

BD = 60√3

Distance between the ships = BD - BC

= 60 - 60√3

= 60 ( 1.732 - 1)

= 43.92

Answer: The distance between the two ships is 43.92cm