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Question
The angle of elevation of the top of an unfinished tower at a point 150 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°?
Solution
Let BC be the length of unfinished tower. Let the tower be raised upto point A so that the angle of elevation at point A is 60°. D is the point on ground from where elevation angles are measured.
In ΔBCD
`"BC"/"CD" = tan 30^circ`
`"BC"/"CD" = 1/sqrt(3)`
`"BC" = "CD"/sqrt(3)`
`"BC" = 150/sqrt(3)` ... (1)
In ΔACD
`("AB + BC")/("CD") = tan 60^circ`
⇒ `("AB + BC")/("CD") = sqrt(3)`
⇒ `("AB" + 150/sqrt(3))/150 = sqrt(3)` ....(Using (i))
⇒ `"AB" = 150sqrt(3) - 150/sqrt(3) = (150 xx 3 - 150)/sqrt(3)`
⇒ `"AB" = 300/sqrt(3) = 300/1.732 = 173.2` m
Thus , the required height is 300 m.
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