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Question
A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower. [Take `sqrt(3)` =1.732 ]
Solution
Let AB be the tower standing vertically on the ground and O be the position of the obsrever we now have:
OA = 20m, ∠OAB = 90° and ∠AOB = 60°
Let
AB = hm
Now, in the right ΔOAB,we have:
`(AB)/(OA)- tan 60° = sqrt(3)`
`⇒ h/20 = sqrt(3)`
`⇒ h = 20 sqrt(3) = (20 xx 1.732 ) = 36.64`
Hence, the height of the pole is 34.64 m.
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