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 30o.Find the height of the tower.​

Answer 1

✬ 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

$\implies{\rm }$ tan30 = AB/BC

$\implies{\rm }$ 1/√3 = x/30

$\implies{\rm }$ 30 = √3x

$\implies{\rm }$ 30/√3 = x

Rationalising the denominator

$\implies{\rm }$ 30/√3 × √3/√3 = x

$\implies{\rm }$ 30√3/3 = x

$\implies{\rm }$ 10√3 = x

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)

Answer 2

Answer:

✬ Height of Tower = 10√3 m ✬

Step-by-step explanation:

Given:

Angle of elevation of top of tower is 30°.

Distance of point from the foot of tower is 30 m.

To Find:

What is the height of tower i.e AB ?

Solution: Let the tower be AB and the point which is 30 m away from foot of tower be C.

➯ Distance of point C from the foot of tower = 30 m

➯ So, BC = 30 m.

Since, the angle of elevation is 30° and the tower is vertical, therefore

➼ ∠ACB = 30° & ∠ABC = 90°.

Now, In right ∆ABC,

tan C = Side opposite to angle C/ Side adjacent to angle C

⟹ tan 30° = AB/BC

⟹ 1/√3 = AB/30

⟹ 30 = √3AB

⟹ 30√3 = AB

• Rationalising the fraction •

⟹ 30√3 \times× √3/√3 = AB

⟹ 30√3/√9 = AB

⟹ 30√3/3 = AB

⟹ 10√3 = AB

Hence, the height of tower AB is 10√3m.

☃️Hence done !!!