Question

25) The angle of elevation of the top of a tower from a point on the ground, which
is30 m away from the foot of the tower, is 30'. Find the height of the tower.
​

Answer 1

Hyy Dude

17.320 m

Step-by-step explanation:

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

Hope it's helps you

Answer 2

Answer:-

/_ACB=Tan30°

•Tan30°= $\frac{1}{\sqrt{3}}$

→Tan30°=$\frac{AB}{CB}$

→$\frac{1}{\sqrt{3}}$=$\frac{h}{30}$

๛Cross Multiply๛

→30=√3h

→$\frac{30}{\sqrt{3}}$=h

→17.32=h[Approx]

$\rule{200}{2}$