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प्रश्न
The angle of elevation of the top of a building from a point A, on the ground, is 30°. On moving a distance of 24 m towards its base to the point B, the angle of elevation changes to 60°. Find the height of the building and distance of point A from the base of the building. (Take `sqrt3` = 1.73)
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उत्तर
From point A, the angle of elevation = 30°. After moving 24 m toward the building to point B, the angle = 60°.
`sqrt3` = 1.73. Find height (h) and distance of A from the base (x).
Let the height of the building = h and the horizontal distance from the base to A = x.
From A (angle 30°): tan 30° = `h/x = 1/sqrt3 ⇒ h = x/sqrt3`.
From B (distance = x − 24, angle 60°):
`tan 60° = h / (x − 24) = sqrt3`
⇒ h = `sqrt3 (x − 24)` ...[Use tan relations tan30 = `1/sqrt3`, tan60 = sqrt3]
Equate the two expressions for h
`x/sqrt3 = sqrt3 (x − 24)`
⇒ multiply both sides by `sqrt3: x = 3(x − 24)`
⇒ x = 3x − 72
⇒ 2x = 72
⇒ x = 36 m.
Height h = `x/sqrt3 = 36/sqrt3 = 12sqrt3. Using sqrt3`
= 1.73, h
= 12 × 1.73
= 20.76 m.
Height = 20.76 m; Distance of A from base = 36 m.
