UPSC CDS 1 2023 Math Paper

Concept:The angle of depression changes as the car approaches the tower. Using trigonometric ratios, we find the distances from the tower base at $30^{\circ}$ and $45^{\circ}$. Since the car moves at uniform speed, time is proportional to distance, allowing us to compute the remaining time $t$.Explanation:Let the height of the tower be $h$ meters.When the angle of depression is $45^{\circ}$, let the distance from the tower base be $d_{1}$.In $\text{triangle ABD}$, $\tan 45^{\circ} = 1 = \frac{h}{d_{1}} \Rightarrow d_{1} = h$.When the angle of depression is $30^{\circ}$, let the distance from the tower base be $d_{2}$.In $\text{triangle ABC}$, $\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{h}{d_{2}} \Rightarrow d_{2} = h\sqrt{3}$.Distance covered by the car in the first 6 minutes is $d_{2} - d_{1} = h\sqrt{3} - h = h(\sqrt{3} - 1)$.Since the car travels at uniform speed, time taken is directly proportional to distance.Let $t$ be the time to cover the remaining distance $d_{1} = h$.Then $\frac{t}{6} = \frac{h}{h(\sqrt{3} - 1)} = \frac{1}{\sqrt{3} - 1}$.Therefore, $t = \frac{6}{\sqrt{3} - 1}$.Rationalizing: $t = \frac{6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{6(\sqrt{3} + 1)}{3 - 1} = 3(\sqrt{3} + 1)$.Using $\sqrt{3} \approx 1.732$, we get $t = 3(1.732 + 1) = 3 \times 2.732 = 8.196$ minutes.This value lies between $8$ and $8.3$ minutes.Answer:Option B: $8 < t < 8.3$