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प्रश्न
From the top of a 15 m high building, the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower and the distance between tower and the building.
उत्तर
Let AB = tower
CD = building
Such that ∠ACE = 30°
∠ADB = 60°
AE = h m (say)
EB = CD = 15 m
and BD = x m (say)
= CE
Now, In ΔAEC, ∠E = 90° we have
`tan 30"°" = h/x`
⇒ `1/sqrt3 = h/x`
⇒ x = `sqrt3 * h` ...(i)
Again in ΔABD, ∠B = 90° we have
`tan 60"°" = (AB)/(BD)`
⇒ `sqrt3 = (h + 15)/x`
⇒ `sqrt3 * (sqrt3 * h) = h + 15` ...[∵ from (i) x = `sqrt3 .` h]
⇒ 3h − h = 15
⇒ h = 7.5 m
⇒ x = `sqrt3h = sqrt3 xx 7.5 ≈ 1.732 xx 7.5` m
x ≈ 12.99 m
Hence height of the tower = 7.5 + 15 = 22.5 m
and distance between tower and building = 12.99 m
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