Question

Two hoardings of cleanliness are put on a 80m wide road on two poles which

are on opposite sides of road and are of same height. The angle of elevation of both

poles from a point on the road is 60º and 30º respectively. Find the height of poles

and distance of poles from that point.

​

Answer 1

Given:

Two hoardings of cleanliness are put on a 80m wide road on two poles which  are on opposite sides of the road and are of the same height.

The angle of elevation of both  poles from a point on the road is 60º and 30º respectively.

To find:

The height of poles  and distance of poles from that point.

​Solution:

Let's assume,

"AB" & "DE" = "h" → the height of the two poles  

"BE = 80 m" → the distance between the two poles

"∠ACB = 60°" → the angle of elevation of one pole

"∠DCE = 30°" → the angle of elevation of another pole

"BC = x m" → the distance of one of the poles from point C

"CE = (80 - x) m" → the distance of another pole from point C

In Δ ABC, we get

$tan \:60\° = \frac{AB}{BC}$

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

$\implies h = x \: \sqrt{3}$ ...... (i)

In Δ DCE, we get

$tan \:30\° = \frac{DE}{CE}$

$\implies \frac{1}{ \sqrt{3}} = \frac{h}{(80 - x)}$

on substituting from (i), we get

$\implies \frac{1}{ \sqrt{3}} = \frac{x\sqrt{3} }{(80 - x)}$

$\implies 80 - x = x\sqrt{3} \times \sqrt{3}$

$\implies 80 - x = 3x$

$\implies 3x + x = 80$

$\implies 4x = 80$

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

$\implies x = 20\:m$

∴ (80 - x) = 80 - 20 = 60 m

On substituting the value of x in eq. (i),we get

$h = 20 \: \sqrt{3}$

Thus,

The height of the two poles is → $\underline{\bold{ 20\sqrt{3}}}$.

The distance of poles from that point is →  $\underline{\bold{20\:m\: \&\: 60 \:m}}$.

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