A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower ?
A. 14 min. 35 sec. B. 15 min. 49 sec. C. 16 min. 23 sec. D. 18 min. 5 sec. Answer: Option CShow Answer
Solution(By Apex Team)
Let AB be the tower and C and D be the point of positions of two cars
Given $\angle \mathrm{ACB}=45^{\circ} \angle \mathrm{ADB}=30^{\circ}$
Let AB=h ,CD=x and AC=y
In ΔABC
$\begin{array}{l}
\tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{AD}} \\
\Rightarrow 1=\frac{\mathrm{AB}}{\mathrm{AD}} \\
\Rightarrow \mathrm{AB}=\mathrm{AD} \Rightarrow \mathrm{y}=\mathrm{h}
\end{array}$
In ΔABD
$\begin{array}{l}\Rightarrow\frac{1}{\sqrt{3}}=\frac{\mathrm{\text{h}}}{\mathrm{\text{x+y}}}\\
\Rightarrow\mathrm{x}+\mathrm{y}=\sqrt{3}\mathrm{~h}\\
\therefore\mathrm{x}=(\mathrm{x}+\mathrm{y})-\mathrm{y}=(\sqrt{3})\mathrm{h}-\mathrm{h}=\mathrm{h}(\sqrt{3}-1)\end{array}$
So car cover distance $\mathrm{h}(\sqrt{3}-1)$ in 12 seconds
Then car cover distance h in $=\left[\frac{12 \times \mathrm{h}}{\mathrm{h}(\sqrt{3}-1)}\right]=\frac{12}{0.73}=$ 16 min 23 sec.
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