Question

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.
• Class 10th !
• Incorrect answers ❌
• Proper answers ✔️

Answer 1

• The height of the tower = 10√3 m.
• The width of the canal = 10 m.

Given :

• ∠ACB = 60° (From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower.)
• CD = 20 m.
• ∠ADB = 30° (From another point (CD) 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower)

To Find :

• The height of the tower.
• The width of the canal.

Solution :

Let,

The height of the tower (AB) be h meters.

The width of the canal (BC) be x meters.

In ∆ABC,

$\bf \implies tan \: {60}^{\circ} = \dfrac{AB}{BC}$

We know that,

• tan 60° = √3.

$\bf \implies \sqrt{3} = \dfrac{h}{x} \\ \\ \\ \bf \implies \sqrt{3} \: x = h \\ \\ \\ \bf \implies h = \sqrt{3} \: x \: \: .........(1)$

In ∆ABD,

$\bf \implies tan \: {30}^{\circ} = \dfrac{AB}{BD}$

We know that,

• tan 30° = 1 / √3.

$\bf \implies \dfrac{1}{ \sqrt{3} } = \dfrac{h}{BC + CD} \\ \\ \\ \bf \implies \dfrac{1}{ \sqrt{3} } = \dfrac{h}{x + 20} \\ \\ \\ \bf \implies x + 20 = \sqrt{3} \: h \: \: ..........(2)$

Now, substitute the equation (1) in the equation (2),

$\bf \implies x + 20 = \sqrt{3} \: h \\ \\ \\\bf \implies x + 20 = \sqrt{3} \: (\sqrt{3} \: x) \\ \\ \\\bf \implies x + 20 = 3x \\ \\ \\\bf \implies 20 = 3x - x \\ \\ \\\bf \implies 20 = 2x \\ \\ \\\bf\implies\dfrac{\not20}{\not2} = x \\ \\ \\\bf \implies 10 = x \\ \\ \\ \: \: \bf \therefore \: \: \: x = 10 \: m$

Therefore, the width of the canal is 10 m.

Now, we have to find the height of the tower.

So, from equation (1),

$\bf \implies h = \sqrt{3} \: x \\ \\ \\ \bf \implies h = \sqrt{3} \: (10 \: m) \\ \\ \\\bf \implies h = \sqrt{3} \times 10 \: m \\ \\ \\ \bf \implies h = 10 \sqrt{3} \: m$

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

Hence,

The height of the tower is 10√3 m and the width of the canal is 10 m.

Answer 2

Answer:

Given :

• the angle of elevation of the top of the tower is 60°.

• From another point 20 m away from this point on the line joining this point to the foot of the tower.

• the angle of elevation of the top of the tower is 30°

To find :

• Find the height of the tower and the width of the canal.

Solution :

Let the height of the tower AB be x metres .

Let the width of the canal BC be w meters .

In ∆ ABC

Tan 60° = √3

=> x/w = √3

=> x = w √3 .

In ∆ ABD

x / ( w + 20 ) = 1/√3

=> √3 x = w + 20

Substitute all values :

=> √3 ( w √3 ) = w + 20

=> 3w = w + 20

=> 3w - w = 20

=> 2w = 20

=> w = 20 / 10

=> w = 10 m

x = w √3

= 10√3 m .

The height of the tower is 10√3 m and the width of the canal is 10 m .