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 100 m apart, the height of the light house is
A. 50/√3+1m
B. 50√3-1m
C. 50(√3-1)m
D.50(√3+1)m

Answer 1

Answer:

The height of light house is 50 ( √3 + 1 ) meters  .

Step-by-step explanation:

Given as :

The angles of depression of two ships from the top of a light house are 45° and 30°

The distance between two ships = d = 100 meters

Let The height of light house = h  meters

According to question

From figure

The height of light house = OC = h meters

The distance OA = x m

The distance OB = (x + 100) m

In Δ OAC

Tan angle = $\dfrac{perpendicular}{base}$

i.e Tan 45° = $\dfrac{OC}{OA}$

Or, Tan 45° = $\dfrac{h}{x}$

i.e  1 = $\dfrac{h}{x}$                         (since Tan 45° = 1)

∴   x = h                      ...............1

Again'

In Δ OBC

Tan angle = $\dfrac{perpendicular}{base}$

i.e Tan 30° = $\dfrac{OC}{OB}$

Or, Tan 30° = $\dfrac{h}{100+x}$

Or, 100 + x = √3 h            ........2        (since Tan 30° = $\dfrac{1}{\sqrt{3} }$  )

From eq 1 and eq 2

Put the value of x

i.e  100 + h = √3 h

Or,  √3 h - h = 100

Or, h ( √3 - 1 ) = 100

∴   h = $\dfrac{100}{\sqrt{3}- 1}$

Or, h = $\dfrac{100}{\sqrt{3}- 1}$  × $\dfrac{\sqrt{3} +1}{\sqrt{3} - 1}$

Or, h = $\dfrac{ 100 (\sqrt{3} + 1 )}{2}$

i.e height = 50 ( √3 + 1 ) meters

So, The height of light house = h =  50 ( √3 + 1 ) meters

Hence, The height of light house is 50 ( √3 + 1 ) meters  . Answer