Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 degree and 45 degree respectively. If the lighthouse is 100 m high, the distance between the two ships is:
(Trigonometry)

dsuc.created: 2 years ago | dsuc.updated: 1 year ago

dsuc.updated: 1 year ago

Let us depict a picture according to question

Now, let one ship is x meter away and another ship is y meter away from the lighthouse.

Here, AD = 100 meter = height of lighthouse

$∠$ ABD = 45 $°$ and $∠$ ACD = 30 $°$

From $∆$ ADC.

tanθ = $\frac{A D}{D C}$

= tan 30 $°$ = $\frac{100}{x}$

$= \frac{1}{\sqrt{3}} = \frac{100}{x} ∴ x = 100 \sqrt{3} A g a i n, f r o m ∆ A B D \tan θ = \frac{A D}{B D} = \tan 45 ° = \frac{100}{y} = 1 = \frac{100}{y} ∴ y = 100 ∵ T h e d i s \tan c e b e t w e e n t h e t w o s h i p s w i l l b e = x + y = 100 \sqrt{3} + 100 = 100 (1.73 + 1) = 100 (2.73) = 273 (a n s w e r)$