Watching from the top of a tower,the angles o | vedclass

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(B) Let the height of the tower be $h$. Let $d_A$ and $d_B$ be the distances of houses $A$ and $B$ from the base of the tower,respectively.From the ge

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house $B$ is closer to the tower than house $A$.

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(B) Let the height of the tower be $h$. Let $d_A$ and $d_B$ be the distances of houses $A$ and $B$ from the base of the tower,respectively.From the geometry of the situation,the angle of depression is equal to the angle of elevation from the house to the top of the tower.Thus,$	an(25^{\circ}) = \frac{h}{d_A}$ and $	an(40^{\circ}) = \frac{h}{d_B}$.This implies $d_A = \frac{h}{	an(25^{\circ})}$ and $d_B = \frac{h}{	an(40^{\circ})}$.Since $40^{\circ} > 25^{\circ}$,we know that $	an(40^{\circ}) > 	an(25^{\circ})$.Therefore,$\frac{h}{	an(40^{\circ})} Hence,house $B$ is closer to the tower than house $A$.

Watching from the top of a tower,the angles of depression of two houses $A$ and $B$ are found to be $25^{\circ}$ and $40^{\circ}$ respectively. Then,.......

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