Question

The angle of elevation of a tower from a point on the ground is

30o

. If we move 20m towards the tower, the angle of elevation

from this point is 60o

. Find the height of the tower.​

Answer 1

Answer:

Height of the tower = 10√3 m

Step-by-step explanation:

Given:

• Angle of elevation of a tower from a point on the ground is 30°
• If we move 20 m towards the tower, the angle of elevation changes to 60°

To Find:

• Height of the tower

Solution:

Let height of the tower be DC.

Let AB = distance travelled = 20 m

Consider Δ BCD

tan 60 = DC/BC

√3 = DC/BC

Cross multiplying,

DC = BC√3-------(1)

Now consider Δ ACD,

tan 30 = DC/AC

tan 30 = DC/(AB + BC)

tan 30 = DC/(20 + BC)

1/√3 = DC/(20 + BC)

Cross multiplying,

DC√3 = 20 + BC

DC = (20 + BC)/√3----(2)

Equating equations 1 and 2,

BC√3 = (20 + BC)/√3

3 BC = 20 + BC

3 BC - BC = 20

2 BC = 20

BC = 10 m

Now substitute the value of BC in equation 1,

DC = 10 × √3

DC = 10√3 m

Hence height of the tower is 10√3 m.

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Answer 2

Answer:

height of the tower is 10√3 m.

Step-by-step explanation:

Given:

Angle of elevation of a tower from a point on the ground is 30°

If we move 20 m towards the tower, the angle of elevation changes to 60°

To Find:

Height of the tower

Solution:

Let height of the tower be DC.

Let AB = distance travelled = 20 m

Consider Δ BCD

tan 60 = DC/BC

√3 = DC/BC

Cross multiplying,

DC = BC√3-------(1)

Now consider Δ ACD,

tan 30 = DC/AC

tan 30 = DC/(AB + BC)

tan 30 = DC/(20 + BC)

1/√3 = DC/(20 + BC)

Cross multiplying,

DC√3 = 20 + BC

DC = (20 + BC)/√3----(2)

Equating equations 1 and 2,

BC√3 = (20 + BC)/√3

3 BC = 20 + BC

3 BC - BC = 20

2 BC = 20

BC = 10 m

Now substitute the value of BC in equation 1,

DC = 10 × √3

DC = 10√3 m

Hence height of the tower is 10√3 m.