Question

answer this question

Answer 1

In the given figure,

AB and CD are two poles of equal height(say h). From a point E on the ground, the angles of elevation of the tops of the AB and CD are 60° and 30° respectively. The distance between the two poles is 80m.

In triangle ABE,

$\tan(60°) = \frac{h}{BE} \\ BE \sqrt{3} = h$

In triangle ECD,

$\tan(30°) = \frac{h}{EC} \\ \frac{EC}{ \sqrt{ 3} } = h$

Equating the above two equations,

$BE \sqrt{3} = \frac{EC}{ \sqrt{3} } \\ 3BE = EC \\ 3BE = 80 - BE \\ 4BE = 80 \\ BE = 20$
So the distance of Pole AB from the point is 20m.

The distance of pole CD from the point is 80-20=60m.

The height of each pole:

$\frac{EC}{ \sqrt{3} } = h \\ \frac{60}{ \sqrt{3} } = h \\ 20 \sqrt{ 3} = h$

Thus the height of each pole is
$20 \sqrt{3} m$