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
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