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Question
From a lighthouse, the angles of depression of two ships on opposite sides of the lighthouse were observed to be 30° and 45°. If the height of the lighthouse is 90 metres and the line joining the two ships passes through the foot of the lighthouse, find the distance between the two ships, correct to two decimal places.
Solution
Let AB is the lighthouse, C and D are the position of two ships.
From right-angled ΔABC,
tan 30° = `"AC"/"BC"`
⇒ `1/sqrt3 = (90 m)/"BC"`
⇒ BC = [ 90 x √3 ] m
∴ BC = 155.88 m
Again, from right angled ΔACD,
tan 45° = `"AC"/"CD"`
⇒ 1 = `(90 m)/"CD"`
⇒ CD = 90 m
Hence, the distance between the two ships
= BC + Cd
= 155.88 + 90 m
= 245.88 m
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