Question

Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet 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:

${ \huge{ \underline { \sf{ \red{Solution:}}}}}$

⇒ Let the width of the road (AB) = 120 feets

⇒ Heights of the poles are AE = BD = hm

⇒ Let 'C' be any point on AB such that from point C

⇒ Angle of elevations are BCD = 60° ; ACE = 30°

⇒ Let BC = x then AC = 120 - x

From right angled Δ CAE :

${ \implies{ \sf{Tan30° = \frac{AE}{AC} }}}$

${ \implies{ \sf{ \frac{1}{ \sqrt{3} } = \frac{h}{120 - x} }}}$

${ \implies{ \sf{120 - x = h \sqrt{3} .....(1)}}}$

From right angled Δ BCD :

${ \implies{ \sf{Tan 60° = \frac{h}{x} }}}$

${ \implies{ \sf{ \sqrt{3} = \frac{h}{x} }}}$

${ \implies{ \sf{h = \sqrt{3}x.....(2) }}}$

From (1) & (2) :

${ \to{ \sf{120 - x = h \sqrt{3} }}}$

${ \to{ \sf{120 - x = ( \sqrt{3}x) \sqrt{3} }}}$

${ \to{ \sf{3x + x = 120}}}$

${ \to{ \sf{4x = 120}}}$

${ \to{ \sf{x = 30}}}$

${ \therefore{ \sf{AC = 120 - x = 120 - 30 = 90 \: feet}}}$

${ \therefore{ \sf{h = \sqrt{3} x = 30 \sqrt{3} = 51.96 \: feet }}}$

Therefore,

• Height of the poles is 30√3 feet

• The distances of the point C from the poles are 30 feet & 90 feet

Answer 2

Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet 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?