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Question
In the given figure, AB is tower of height 50 m. A man standing on its top, observes two car on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the two cars.
Solution
Clearly ∠ADB = 45° and ∠ACB = 30° ....[∵ Angle of depression = Angle of elevation]
Now, In ΔABD, we have
tan 45° = `(AB)/(BD)`
⇒ 1 = `50/(BD)`
⇒ BD = 50 m ...(i)
Also, In ΔABC, we have
tan 30° = `(AB)/(BC)`
⇒ `1/sqrt(3) = (AB)/(BC)`
⇒ BC = `ABsqrt(3) = 50sqrt(3)` m ...(ii)
From equations (i) and (ii), we get
CD = BC + BD
= `(50sqrt(3) + 50)`m
= `50(sqrt(3) + 1)`m
= 50(1.732 + 1)
= 50 × 2.732
= 136.6 m
Thus, the distance between two cars is 136.6 m.
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