Question

309) two poles of equal tieghts are stand
Ta opposite to each other on either side
of the road which is 80 mts wide
From a point between then on the road the angles of elevation of the top of the poles are 60 degrees and 30 degrees respectively find the height of the pole

Answer 1

Answer :-

Height of the pole is 34.64 m.

Explanation :-

According to the given information draw a figure

[ Refer the attachment ]

Let the to poles be be AB and DE

Height of AB = Height of CD

Distance between AB and CD = 80 m

Let the point C be at a distance of 'x' m from AB

Angle of elevation to the top of the pole AB to A ∠ACB = 60°

Angle of elevation to the top of the pole DE to E ∠ECD = 30°

From figure,

→ BD = 80 m

→ BC = 'x' m

→ CD = BD - BC = ( 80 - x ) m

Consider ΔABC

tan 60° = Opposite side / Adjacent side

⇒ √3 = AB / BC

[ ∵ tan 60° = √3 ]

⇒ √3 = AB / x

⇒ √3 * x = AB

⇒ x = AB / √3 → eq ( 1 )

Consider ΔCDE

tan 30° = Opposite side / Adjacent side

⇒ 1 / √3 = DE / CD

[ ∵ tan 30° = 1 / √3 ]

⇒ 1 / √3 = DE / ( 80 - x )

⇒ 80 - x = √3 * DE

⇒ 80 - √3 * DE = x

⇒ x = 80 - √3 * DE

⇒ x = 80 - √3 * AB → eq ( 2 )

[ ∵ DE = AB ]

From eq ( 1 ) and eq ( 2 )

⇒ AB / √3 = 80 - √3 * AB

⇒ AB = √3 ( 80 - √3 * AB )

⇒ AB = 80√3 - 3AB

⇒ AB + 3AB = 80√3

⇒ 4AB = 80√3

⇒ AB = 80√3 / 4

⇒ AB = 20√3

⇒ AB = 20 * 1.732

[ ∵ √3 ≈ 1.732 ]

⇒ AB = 34.64

Hence, height of the pole is 34.64 m.

Answer 2

SOLUTION:-

Given:

•Two poles of equal height are standing opposite to each other on either side of the road, which is 80m wide.

•From a point between them on the road, the angles of elevation of the top of the poles are 60° & 30° respectively.

To find:

The height of the pole.

Explanation:

Let the poles be AB & DE, & C be the point of observation.

We have,

•Width of the road, BD= 80m

•Angle of elevation to AB, ∠ACB=30° & •Angle of elevation to DE, ∠ECD=60°

In right angled ∆ACB,

$tan \theta = \frac{Perpendicular}{Base}$

$tan30 \degree = \frac{AB}{BC} \\ \\ \frac{1}{ \sqrt{3} } = \frac{AB}{BC} \\ \\ \sqrt{3} AB = BC \\ \\ AB = \frac{BC}{ \sqrt{3} } ...............(1)$

&

In right angled ∆EDC,

$tan60 \degree = \frac{ED}{CD} \\ \\ \sqrt{3} = \frac{ED}{80 - BC} \\ \\ ED = \sqrt{3} (80 - BC)..............(2)$

We know that,AB= ED as the poles are of same height.

Therefore,

Comparing equation (1) & (2), we get;

$\frac{BC}{ \sqrt{3} } = \sqrt{3} (80 - BC) \\ \\ BC = 3(80 - BC) \\ \\ BC = 240 - 3BC \\ \\ BC + 3BC = 240 \\ \\ 4BC= 240 \\ \\ BC = \frac{240}{4} \\ \\ BC = 60m$

Now,

Using the value of BC in equation (1), we get;

&

AB is the height of the pole:

$AB = \frac{60}{ \sqrt{3} } \\ [Rationalise] \\ \\ AB = \frac{60 \sqrt{ 3} }{ \sqrt{3} \times \sqrt{3} } \\ \\ AB = \frac{60 \sqrt{3} }{3} m \\ \\ AB = 20 \sqrt{3} m$

Or

AB= (20× 1.732)m [√3= 1.732]

AB= 34.64m.