On the level ground,the angle of elevation of | vedclass

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(B) Let the height of the tower be $h$ and the distance from the tower to the second point be $x$.In $	riangle BAQ$,$	an 60^{\circ} = \frac{h}{x} \i

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$10 \sqrt{3}$

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(B) Let the height of the tower be $h$ and the distance from the tower to the second point be $x$.In $	riangle BAQ$,$	an 60^{\circ} = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \dots(1)$In $	riangle BAP$,$	an 30^{\circ} = \frac{h}{x+20} \implies \frac{1}{\sqrt{3}} = \frac{h}{x+20} \implies x+20 = h\sqrt{3} \quad \dots(2)$Substitute $x = \frac{h}{\sqrt{3}}$ from $(1)$ into $(2)$:$\frac{h}{\sqrt{3}} + 20 = h\sqrt{3}$$20 = h\sqrt{3} - \frac{h}{\sqrt{3}}$$20 = h \left( \frac{3-1}{\sqrt{3}} ight)$$20 = h \left( \frac{2}{\sqrt{3}} ight)$$h = \frac{20 	imes \sqrt{3}}{2} = 10\sqrt{3} \ m$.

On the level ground,the angle of elevation of the top of the tower is $30^{\circ}$. On moving $20 \ m$ nearer,the angle of elevation is $60^{\circ}$. The height of the tower is (in $m$):

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