The angles of elevation of the top of a vertical tower from two points. 30 metros apart, and on the same straight line passing through the base of tower, are 300 and 600 respectively. The height of the tower is
A
10 m
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B
15 m
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C
15√3
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D
30 m
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Solution
The correct option is C15√3 Let AB is tower, the angles of elevation of the top of a vertical tower AB from two points D and C are 300 and 600 respectively. Let the height of tower be h metres and BC=xm In triangle ABC,ABBC=tan600 ∴hx=√3...........(i) In triangle ABD,ABBD=tan300 ∴h(x+30)=1√3 or √3h=x+30 Put value of x from equation Ci) in equation (ii), we get √3h=h√3+30 or √3h−h√3=30 or 2h√3=30 ∴h=30√32=15√3 So, height of the tower is 15√3metre.