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

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.

Answer 2

• Let AB = CD = h = height of pole.

BE = x

BD = 80 m

ED = (80 - x) m = distance of the point from the poles.

In ∆ABE

→ tan60° = $\dfrac{AB}{BE}$

→ √3 = $\dfrac{h}{x}$

→ x√3 = h

→ h = x√3 _______ (eq 1)

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In ∆CED

→ tan30° = $\dfrac{CD}{DE}$

→ $\dfrac{1}{\sqrt{3}}$ = $\dfrac{h}{80\:-\:x}$

→ h√3 = 80 - x

→ (x√3)√3 = 80 - x [From (eq 1)]

→ 3x = 80 - x

→ 2x = 80

→ x = 60 m

→ ED = 80 - 60

→ ED = 20 m

Put value of x in (eq 1)

→ h = 20√3 m

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Height of pole = 20√3 m and BE = 60 m and ED = 20 m (distance of the poles).

_______ [ ANSWER ]

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