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:

Let AB and DE are two poles and given AB = DE

Let BD = 80 is the road.

Let there is a point C on the road.

Let CD = x then BC = 80 - x

Now from triangle EDC,

tan 30 = ED/CD

=> 1/√3 = ED/x

=> x = ED√3 ........1

Again from triangle ABC

tan60 = AB/BC

=> √3 = AB/(80 - x)

=> AB = √3(80 - x)

=> AB = 80√3 - √3x

=> AB = 80√3 - √3*ED*√3 ( x = ED√3)

=> AB = 80√3 - 3ED

=> AB = 80√3 - 3AB (ED = AB)

=> AB + 3AB = 80√3

=> 4AB = 80√3

=> AB = 80√3/4

=> AB = 20√3

So AB = ED = 20√3

So AB = ED = 20√3So height of the pole is 20√3 m.

Now from equation 1

x = ED√3

=> x = 20√3*√3

=> x = 20*3

=> x = 60

So CD = 60

and BC = 80 - x = 80 - 60

=> BC = 20

So distances of the point from the poles are 60 m and 20 m.

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Answer 2

Answer:

Solution

The values we have,

AB = 80 cm

AC = BD = h

AP = x and BP = 80 - x

→ Perpendicular/Base = tan∅

→ AC/AP = tan 30°

→ AC/x = 1/√3

→ AC = x/√3

→ BD/BP = tan 60°

→ BD/(80 - x) = √3

→ BD = √3(80 - x)

Since, AC = BD

→ x/√3 = √3(80 - x)

→ x = 240 - 3x

→ 4x = 240

→AP = x = 60

So, BP = (80 - 60) = 20

→ AC = 60/√3

→ AC = 60√3/3

→ AC = BD = 20√3

Answer

Height of both poles is 20√3 m and point is 60 m away from left pole and 20 m away from right pole.