As observed from the top of a 100 m high light house

Trigonometry (10)

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 $\sqrt{3}$ = 1.732]

Answer

Let AB be the tower and ships are at points C and D. 

$$ \tan45^0 = \frac{AB}{BC} $$

$$ \frac{AB}{BC} = 1 $$

AB = BC 

$$ 30^0 = \frac{1}{\sqrt{3}} = \frac{AB}{BC+CD} $$

$$ \frac{1}{\sqrt{3}} = \frac{AB}{BC+CD} $$

$$ AB +CD= \sqrt{3}AB $$ 

$$ CD = AB(\sqrt{3} - 1) $$

= 100 x (1.732 - 1)

= 73.2 m

• Exam Year: 2018