The angle of a elevation to the top of a tower from a point 30m away from the foot of tower on the ground is 30.find the height of tower.
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Answer:
17.320 m
Refer the attached figure
AB = Height of tower
BC = 30 m
∠ACB = 30°
InΔABC
Tan \theta=\frac{Perpendicular}{Base}
Tan30^{\circ}=\frac{AB}{BC}
30 \times \frac{1}{\sqrt{3}}=AB
17.320=AB
Hence The height of tower is 17.320 m
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