Question

a man standing a top of a light house sees a ship at angle of depression 22 after the ship approaching towards the seashore 100metres then he sees at an angle of depression 31?find the height of light house? how far the ship from lighthouse ?​

Answer 1

Given:

A man standing a top of a lighthouse sees a ship at an angle of depression 22°

after the ship approaching towards the seashore 100metres then he sees at an angle of depression 31°

To find:

The height of the lighthouse and how far the ship is from the lighthouse

Solution:

​To solve the given problem we will use the following formula of trigonometric ratios:

$\boxed{\bold{tan\:\theta = \frac{Perpendicular}{Base} }}$

 

Referring to the figure, we have

"AB" → height of the lighthouse

"CD" → distance moved by ship towards the seashore = 100 m

"∠XAD" = 22° → angle of depression of the ship

"∠XAC" = 31° → angle of depression of the ship after approaching towards the seashore

∴ ∠XAD = ∠ADC = 22° and ∠XAC = ∠ACB = 31°  ...... (alternate angles)

Considering Δ ABC, we have

θ = 31°

Perpendicular = AB

Base = BC

∴  $tan \:31\° = \frac{AB}{BC}$

$\implies 0.6 = \frac{AB}{BC}$

$\implies AB = 0.6 BC$ ....... (i)

Considering Δ ABD, we have

θ = 22°

Perpendicular = AB

Base = BC + CD

∴  $tan \:22\° = \frac{AB}{BC + CD}$

$\implies 0.404 = \frac{AB}{BC + 100}$

$\implies AB = 0.404BC + 40.40$ ....... (ii)

Comparing equations (i) & (ii), we get

 $0.6 BC = 0.404BC + 40.40$

$\implies 0.6 BC - 0.404BC = 40.40$

$\implies 0.196 BC = 40.40$

$\implies BC = \frac{40.40}{0.196}$

$\implies BC = 206.12\:m$

 

On substituting the value of BC in eq. (i), we get,

$\implies AB = 0.6 \times 206.12 = 123.67\:m$

 

Thus,

$\boxed{\bold{The \:height\:of\:the\:lighthouse\:is\:\underline{123.67\:m}}}.\\\\\boxed{\bold{The \:ship\:is\:\underline{206.12\:m}\:far\:from\:the\:lighthouse}}.$

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