The angle of elevation of the top of tower N | vedclass

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(C) Let the height of tower $M$ be $h_M$ and the height of tower $N$ be $h_N$. Let the distance between the bases of the two towers be $d$.From the ba

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tower $M$ is taller than tower $N$.

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(C) Let the height of tower $M$ be $h_M$ and the height of tower $N$ be $h_N$. Let the distance between the bases of the two towers be $d$.From the base of tower $M$,the angle of elevation to the top of tower $N$ is $40^{\circ}$,so $	an(40^{\circ}) = \frac{h_N}{d}$,which implies $h_N = d \cdot 	an(40^{\circ})$.From the base of tower $N$,the angle of elevation to the top of tower $M$ is $55^{\circ}$,so $	an(55^{\circ}) = \frac{h_M}{d}$,which implies $h_M = d \cdot 	an(55^{\circ})$.Since $55^{\circ} > 40^{\circ}$,we know that $	an(55^{\circ}) > 	an(40^{\circ})$.Therefore,$h_M > h_N$,which means tower $M$ is taller than tower $N$.

The angle of elevation of the top of tower $N$ from the base of tower $M$ is $40^{\circ}$. The angle of elevation of the top of tower $M$ from the base of tower $N$ is $55^{\circ}$. Then,.......

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