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(B) Let the height of the tower be $h$ and the foot of the tower be $O$.Let $x_A$ be the distance of house $A$ from the tower and $x_B$ be the distanc

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House $B$ is nearer to the tower than House $A$

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(B) Let the height of the tower be $h$ and the foot of the tower be $O$.Let $x_A$ be the distance of house $A$ from the tower and $x_B$ be the distance of house $B$ from the tower.From the geometry of the problem,the angle of depression equals the angle of elevation.For house $A$: $	an(30^{\circ}) = \frac{h}{x_A} \implies \frac{1}{\sqrt{3}} = \frac{h}{x_A} \implies x_A = h\sqrt{3}$.For house $B$: $	an(60^{\circ}) = \frac{h}{x_B} \implies \sqrt{3} = \frac{h}{x_B} \implies x_B = \frac{h}{\sqrt{3}}$.Comparing the distances,since $\sqrt{3} > \frac{1}{\sqrt{3}}$,it follows that $x_A > x_B$.Therefore,house $B$ is closer to the tower than house $A$.

When observed from the top of a tower,the angles of depression of two houses $A$ and $B$ in the Eastern and Western directions are $30^{\circ}$ and $60^{\circ}$ respectively. Then........

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