The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower...
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Answer:
AB = Height of tower
BC = 30 m
∠ACB = 30°
InΔABC
Tan \theta=\frac{Perpendicular}{Base}Tanθ=
Base
Perpendicular
Tan30^{\circ}=\frac{AB}{BC}Tan30
∘
=
BC
AB
30 \times \frac{1}{\sqrt{3}}=AB30×
3
1
=AB
17.320=AB17.320=AB
Hence The height of tower is 17.320
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