Question

The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 200 m apart, find the height of the light house.

Answer 1

Given: From the top of a light house the angles of depression to two ships both towards east are 45° and 30° and the ships are 200 m apart.

Step-wise calculation:

1. Let the height of the light house be h. Let horizontal distances from the base of the light house to the ship with depression 45° and to the ship with depression 30° be x₁ and x₂ respectively.

2. Using tan of angle of depression equal to tan of corresponding angle of elevation:

`tan 45^circ = h/x_1`

⇒ `1 = h/x_1`

⇒ x₁ = h

`tan 30^circ = h/x_2`

⇒ `1/sqrt(3) = h/x_2`

⇒ `x_2 = hsqrt(3)`

3. The ships are on the same side east, so their horizontal separation is x₂ – x₁ = 200.

`x_2 - x_1 = hsqrt(3) - h`

= `h(sqrt(3) - 1)`

= 200

4. Solve for h:

`h = 200/(sqrt(3) - 1)`

Rationalize:

`h = (200(sqrt(3) + 1))/(3 - 1)` 

= `100(sqrt(3) + 1)`

5. Numerical value:

`sqrt(3) ≈ 1.7320508`

So, h = 100 × 2.7320508

= 273.20508

= 273.21 m

The height of the light house is `100(sqrt(3) + 1)` metres, which is approximately 273.21 m.