Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are $60^{\circ} a n d 30^{\circ}$ , find the height of the pillars and the position of the point.

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Solution

Let the height of the equal pillars be AB = CD = h

Given, width of the road is 150 m

Let BE = x, the DE = 150 - x

In right angle triangle ABE,

tan 60 = $\frac{h}{x}$

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

=> $h = \sqrt{3} x$ ............1

In right angle triangle CDE,

$tan 30 = \frac{h}{(150 - x)}$

=> $\sqrt{3} h = 150 - x$

=> $\sqrt{3} h = 150 - \frac{h}{\sqrt{3}}$ {from equation 1}

=> $h = \frac{150 \sqrt{3}}{4}$

=> $h = 37.5 \sqrt{3} m$

So, the height of the equal pillars is $37.5 \sqrt{3} m$

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Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are $60^{\circ} a n d 30^{\circ}$ , find the height of the pillars and the position of the point.