Question

The angle of elevation of the top of a tower from a point on the ground, which is 30m
away from the foot of the tower is 30°. The height of tower is
​

Answer 1

Answer:

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

Answer 2

Solution :

In ∆ ABC,

tan ∅ = Opposite side / Adjacent side

tan 30° = AB/BC

1/√3 = AB/30

30/√3 = AB

AB = 30/√3

Now, Multiplying numerator and denominator by 3 we get:

AB = 30/√3 × √3/√3

AB = 30√3/3

AB = 10√3

Therefore, the height of the tower is 10√3.