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(A) Let the height of each pillar be $h$ and the width of the road be $60 \, m$. Let the point on the road be at a distance $x$ from one pillar,so the

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$15\sqrt{3} \, m$

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(A) Let the height of each pillar be $h$ and the width of the road be $60 \, m$. Let the point on the road be at a distance $x$ from one pillar,so the distance from the other pillar is $(60 - x)$.In the first triangle,$	an 60^\circ = \frac{h}{x}$ $\Rightarrow \sqrt{3} = \frac{h}{x}$ $\Rightarrow h = x\sqrt{3} \dots (i)$.In the second triangle,$	an 30^\circ = \frac{h}{60 - x}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{60 - x}$ $\Rightarrow 60 - x = h\sqrt{3} \dots (ii)$.Substituting $x = \frac{h}{\sqrt{3}}$ from $(i)$ into $(ii)$:$60 - \frac{h}{\sqrt{3}} = h\sqrt{3}$$60 = h\sqrt{3} + \frac{h}{\sqrt{3}} = h \left( \frac{3+1}{\sqrt{3}} ight) = \frac{4h}{\sqrt{3}}$$h = \frac{60 	imes \sqrt{3}}{4} = 15\sqrt{3} \, m$.

Two pillars of equal height stand on either side of a roadway which is $60 \, m$ wide. At a point in the roadway between the pillars,the angles of elevation of the top of the pillars are $60^\circ$ and $30^\circ$. The height of the pillars is

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