Question

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...​

Answer 1

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

Answer 2

Answer:

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