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

$\large\underline{\underline{\rm{\red{Given:-}}}}$

$\rm \angle APB = 60^0$

$\rm \angle DPC = 30^0$

$\rm BC = 80 \ m$

$\large\underline{\underline{\rm{\red{Solution:-}}}}$

Let BP = x , CP = 80 - x

Now , in $\rm \triangle ABP$

$\rm :\longmapsto tan60^0 = \dfrac{AB}{BP}$

$\rm :\longmapsto \sqrt3 = \dfrac{AB}{x}$

$\rm :\longmapsto AB = \sqrt3x$

Now in $\rm \triangle BPC$

$\rm :\longmapsto tan30^0 = \dfrac{CD}{CP}$

$\rm :\longmapsto \dfrac{1}{\sqrt3} = \dfrac{CD}{80 - x}$

$\rm :\longmapsto CD = \dfrac{80-x}{\sqrt3}$

As AB = CD,

$\rm :\longmapsto \sqrt3 x = \dfrac{80-x}{\sqrt3}$

$\rm :\longmapsto 3x = 80 - x$

$\rm :\longmapsto 4x = 80$

$\rm :\longmapsto x = 20\ m$

Hence, height of pole is given as:

$\boxed{\boxed{\rm h = \sqrt3x = 20\sqrt3\ m}}$

Distance of point from pole AB

$\boxed{\boxed{\rm BP = x \ m = 20 \ m}}$

Distance of point from pole CD

[tex]\boxed{\boxed{\rm CP = 80 - x = 60 \ m}}