Question

Two poles of equal heights are standing opposite each other on either side of the roads, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.​

Answer 1

Answer:

Given AB = 80 m

Let AP = x m, therefore, PB = (80 - x) m

In APC,

tan30° = AC/AP

1/√3 = h/x ......(1)

In BPD,

tan60° = BD/AB

√3 = h/(80 - x) ......(2)

Dividing eqn (1) by (2) , we get

1/√3/√3 = h/x/h/(80 - x)

=> 1/3 = (80 - x)/x

=> x = 240 - 3x

=> 4x = 240

=> x = 60

from eqn (1),

1/√3 = h/x

=> h = 60/√3

=> h = 20√3 m

Thus, the height of both the poles is 20√3 m and the distances of the point from the poles are 60 m and 20 m.

Answer 2

SOLUTION

Let AB and DE be the two poles, & C be the points of observation.

Given,

Width of road, BD=80m,

Angle of elevation to AB,

angle ACB= 30°

Angle of elevation to DE, angle ECD=60°

To find

Height of buildings AB & DE.

In ∆ACB

$tan \theta = \frac{P}{B} \\ tan30 \degree = \frac{AB}{BC} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{AB}{BC} \\ \\ = > AB = \frac{BC}{ \sqrt{3} } ..........(1)$

In ∆EDC,

$tan \theta = \frac{P}{B} \\ \\ tan60 \degree = \frac{ED}{CD} \\ \\ = > \sqrt{3} = \frac{ED}{80 - BC} \\ \\ = > ED = \sqrt{3} (80 - BC)..........(2)$

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

Hence, from (1) & (2),

$\frac{</u><u>BC</u><u>}{ \sqrt{3} } = \sqrt{3} (80 - BC ) \\ </u><u>C</u><u>ross \: multiplying \: we \: get \\ = > bc = 3(80 - </u><u>BC</u><u>) \\ = > </u><u>BC</u><u> = 240 - 3</u><u>B</u><u>C</u><u> \\ = > 4</u><u>B</u><u>C</u><u> = 240 \\ = > </u><u>BC</u><u> = \frac{240}{4} \\ = > </u><u>BC</u><u> = 60m$

Now, using the value of BC in (2),

AB= 20√3m

$tan60 \degree = \frac{ED}{CD}$

Now from ∆EDC,

$= > \sqrt{3} = \frac{20 \sqrt{3} }{CD} \\ \\ = > CD = 20m$

Therefore, Height of pole = 20√3m

Distance of poles from observing point

=) 60m & 20m

Hope it helps ☺️