Question

The angle of elevation of the top of a tower from a point on ground which is 30 meters away from the foot of the tower is 30°. Height of the tower is?

Answer 1

Step-by-step explanation:

In ∆ ABC,

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.

Answer 2

✬ Height = 17.32 m or 10√3 ✬

Step-by-step explanation:

Given:

Angle of elevation from the top of tower is 30°.

Distance between foot of tower and point on ground us 30 m.

To Find:

What is the height of tower ?

Solution:

Let AB be a tower of height x m and BC be the distance between foot of tower and point on ground.

Now , In ∆ABC we have

AB = Height (Perpendicular) = x m.

BC = Base = 30 m.

∠ACB = 30°

As we know that

$★ tanθ = Perpendicular/Base ★$

Here in this ∆ we have

∠ACB = θ = 30°

$Applying \: tanθ \: in \: ∆ABC \\ </p><p></p><p>\implies{\rm }⟹ tan30 = AB/BC \\ </p><p></p><p>\implies{\rm }⟹ 1/√3 = x/30 \\ </p><p></p><p>\implies{\rm }⟹ 30 = √3x \\ </p><p></p><p>\implies{\rm }⟹ 30/√3 = x \\ </p><p></p><p>Rationalising \: the \: denominator</p><p>$

$\implies{\rm }⟹ 30/√3 × √3/√3 = x \\ </p><p></p><p>\implies{\rm }⟹ 30√3/3 = x \\ </p><p></p><p>\implies{\rm }⟹ 10√3 = x \\ </p><p>$

Hence, the height of tower is 10√3 m.

To find approx height of tower let's put the value of √3 i.e 1.732

Height = 10 × 1.732 = 17.32 m (approx)