At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is 512. On walking 160 metres towards the tower, the tangent of the angle of elevation is 34. The height of the tower is equal to
A
150m
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B
180m
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C
200m
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D
None
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Solution
The correct option is A150m Let h be the height of the tower.
Given, at 1st point on the ground, tanα=512 ...given ∴tanα=hx+160
Now we can write 512=hx+160
12h=5x+800 ...(1) At second point on the ground, tanβ=34 ...given tanβ=hx
where, x is the distance of 2nd point on the ground from tower Now we can write hx=34 ∴x=4h3 ...(2) Put the value of x from eq 2 in eq 1, we get h=150m