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 hight of the tower
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Answer:
17.320
Step-by-step explanation:
Refer the attached figure
AB = Height of tower
BC = 30 m
∠ACB = 30°
InΔABC
Tan \theta=\frac{Perpendicular}{Base}Tanθ=BasePerpendicular
Tan30^{\circ}=\frac{AB}{BC}Tan30∘=BCAB
30 \times \frac{1}{\sqrt{3}}=AB30×31=AB
17.320=AB17.320=AB
Hence The height of tower is 17.320 m
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