The angles of elevation of the top of a tower | vedclass

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Question

(A) Let the height of the tower be $h$ and the foot of the tower be $P$. Let $CP = x$,$BC = y$,and $AB = z$.In $	riangle QCP$,$	an 60^{\circ} = \fra

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$\sqrt{3}: 1$

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(A) Let the height of the tower be $h$ and the foot of the tower be $P$. Let $CP = x$,$BC = y$,and $AB = z$.In $	riangle QCP$,$	an 60^{\circ} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}$.In $	riangle QBP$,$	an 45^{\circ} = \frac{h}{x+y}$ $\Rightarrow x+y = h$ $\Rightarrow y = h - \frac{h}{\sqrt{3}} = h\left(1 - \frac{1}{\sqrt{3}}ight) = h\left(\frac{\sqrt{3}-1}{\sqrt{3}}ight)$.In $	riangle QAP$,$	an 30^{\circ} = \frac{h}{x+y+z}$ $\Rightarrow x+y+z = h\sqrt{3}$ $\Rightarrow z = h\sqrt{3} - (x+y) = h\sqrt{3} - h = h(\sqrt{3}-1)$.Now,the ratio $\frac{AB}{BC} = \frac{z}{y} = \frac{h(\sqrt{3}-1)}{h\left(\frac{\sqrt{3}-1}{\sqrt{3}}ight)} = \sqrt{3}$.Thus,$AB: BC = \sqrt{3}: 1$.

The angles of elevation of the top of a tower from three collinear points $A, B$,and $C$ on a road leading to the foot of the tower are $30^{\circ}, 45^{\circ}$,and $60^{\circ}$ respectively. The ratio of $AB$ to $BC$ is

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