Question

Two poles of equal heights are standing opposite each other on either side of the road, 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

Let ,

• The distance between B and O be " x "

In Δ ABO ,

$\tt \implies \tan(60) = \frac{AB}{x}$

$\tt \implies \sqrt{3}x = AB - - - (i)$

Now , in Δ ADO

$\tt \implies \tan(30) = \frac{CD}{80 - x}$

$\tt \implies\frac{1}{ \sqrt{3} } = \frac{CD}{80 - x}$

$\tt \implies CD = \frac{80 - x}{ \sqrt{3} }$

Since , the height of two poles is same

Thus ,

$\tt \implies \sqrt{3}x = \frac{80 - x}{ \sqrt{3} }$

$\tt \implies 3x = 80 - x$

$\tt \implies 4x = 80$

$\tt \implies x = \frac{80}{4}$

$\tt \implies x = 20 \: \: m$

Therefore , The distance between the points from poles are 20 m and 60 m

Put the value of x = 20 m in eq (i) , we get

$\tt \implies height = 20 \sqrt{3 }$

$\tt \implies height =20 \times 1.732$

$\tt \implies height =34.64 \: \: m$

The height of two poles is 34.64 m

Answer 2

Given :-

Width = 80 m

Angles of the elevation from the top = 60° and 30°

To Find :-

The height of the poles and the distances of the point from the poles.

Solution :-

(Refer to the attachment provided)

Let us consider AB and CD to be the poles of equal height.

O is the point between them from where the height of elevation taken. BD is the distance between the poles.

According to the figure,

$\longrightarrow \sf AB = CD$

$\longrightarrow \sf OB + OD = 80 \: m$

Now,  in right ΔCDO,

$\longrightarrow \sf tan \: 30^{o} = \dfrac{CD}{OD}$

$\longrightarrow \sf \dfrac{1}{\sqrt{3} } =\dfrac{CD}{OD}$

$\implies \sf CD=\dfrac{OD}{\sqrt{3} } \qquad ...(1)$

Again,  in right ΔABO,

$\longrightarrow \sf tan\: 60^{o} =\dfrac{AB}{OB}$

$\longrightarrow \sf \sqrt{3} =\dfrac{AB}{(80-OB)}$

$\implies \sf AB=\sqrt{3} (80-OD)$

AB = CD (Given)

Using equation (1),

$\longrightarrow \sf \sqrt{3} (80-OD) = \dfrac{OD}{\sqrt{3} }$

$\implies \sf 3(80-OD) = OD$

$\implies \sf 240 - 3 \: OD = OD$

$\implies \sf 4 \: OD = 240$

$\implies \sf OD = 60$

Putting the value of OD in equation (1)

$\longrightarrow \sf CD = \dfrac{OD}{\sqrt{3} }$

$\implies \sf CD = \dfrac{60}{\sqrt{3} }$

$\implies \sf CD =20\sqrt{3} \: m$

Also,

$\longrightarrow \sf OB + OD = 80 \: m$

$\implies \sf OB = (80-60) \: m = 20 \: m$

Therefore, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and  60 m respectively.