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(A) Let $h$ be the height of the hill $QP$ and $x$ be the distance from the second point $A$ to the base $P$ of the hill.In $	riangle QPA$,$	an 60^{

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$\frac{3}{2} 	ext{ km}$

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(A) Let $h$ be the height of the hill $QP$ and $x$ be the distance from the second point $A$ to the base $P$ of the hill.In $	riangle QPA$,$	an 60^{\circ} = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies h = \sqrt{3}x$.In $	riangle QPO$,the distance $OA = \sqrt{3} 	ext{ km}$,so $OP = OA + AP = \sqrt{3} + x$.$	an 30^{\circ} = \frac{h}{\sqrt{3} + x} \implies \frac{1}{\sqrt{3}} = \frac{h}{\sqrt{3} + x} \implies \sqrt{3}h = \sqrt{3} + x$.Substitute $h = \sqrt{3}x$ into the equation: $\sqrt{3}(\sqrt{3}x) = \sqrt{3} + x \implies 3x = \sqrt{3} + x \implies 2x = \sqrt{3} \implies x = \frac{\sqrt{3}}{2} 	ext{ km}$.Therefore,$h = \sqrt{3} 	imes \frac{\sqrt{3}}{2} = \frac{3}{2} 	ext{ km}$.

$A$ person walking along a straight road towards a hill observes at two points,separated by a distance of $\sqrt{3} \text{ km}$,that the angles of elevation of the hill are $30^{\circ}$ and $60^{\circ}$. The height of the hill is

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