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(D) Let $H = 150 \, m$ be the height of the second chimney.Let $d$ be the distance between the two chimneys.From the geometry of the problem,the angle

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(D) Let $H = 150 \, m$ be the height of the second chimney.Let $d$ be the distance between the two chimneys.From the geometry of the problem,the angle of depression to the foot of the first chimney is $\phi$,so $	an \phi = \frac{H}{d}$.Given $	an \phi = \frac{5}{2}$,we have $\frac{150}{d} = \frac{5}{2}$,which gives $d = \frac{150 	imes 2}{5} = 60 \, m$.Let $h$ be the height of the first chimney. The angle of depression to the top of the first chimney is $	heta$,so $	an 	heta = \frac{H - h}{d}$.Given $	an 	heta = \frac{4}{3}$,we have $\frac{150 - h}{60} = \frac{4}{3}$.$150 - h = 60 	imes \frac{4}{3} = 80$.$h = 150 - 80 = 70 \, m$.The vertical distance between the tops is $H - h = 150 - 70 = 80 \, m$.The horizontal distance between the tops is $d = 60 \, m$.The distance between the tops is $\sqrt{(H - h)^2 + d^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, m$.

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