# 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 C
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.