A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the Fig. 9.12 Fig. 9.13 143 tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Step-by-Step Solution
Key Concept: Use the definition of tangent for the angle of depression (tan θ = height / horizontal distance). The two observations give two horizontal distances from the foot of the tower. Since the car moves with uniform speed, the distance covered in the known time interval gives the speed, which is then used to find the remaining time to reach the foot.
1. Let the height of the tower be $h$ metres.
2. At the first observation (angle of depression $30^{\circ}$), the horizontal distance of the car from the foot of the tower is
$$d_1 = \frac{h}{\tan 30^{\circ}} = \frac{h}{\frac{1}{\sqrt3}} = h\sqrt3.$$
3. After 6 s the angle of depression becomes $60^{\circ}$. The new horizontal distance is
$$d_2 = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt3} = \frac{h}{\sqrt3}.$$
4. The car travels the distance $d_1-d_2$ in 6 s, so its uniform speed $v$ is
\begin{align*}
v &= \frac{d_1-d_2}{6}
= \frac{h\sqrt3-\frac{h}{\sqrt3}}{6}
= \frac{h\left(\sqrt3-\frac{1}{\sqrt3}\right)}{6}
= \frac{h\left(\frac{3-1}{\sqrt3}\right)}{6}
= \frac{2h}{6\sqrt3}
= \frac{h}{3\sqrt3}\;\text{m/s}.
\end{align*}
5. The remaining distance to the foot of the tower after the second observation is $d_2 = \frac{h}{\sqrt3}$.
6. Time $t$ required to cover this distance at speed $v$ is
\begin{align*}
t &= \frac{d_2}{v}
= \frac{\frac{h}{\sqrt3}}{\frac{h}{3\sqrt3}}
= \frac{h}{\sqrt3}\times\frac{3\sqrt3}{h}
= 3\;\text{seconds}.
\end{align*}
7. Hence, the car will reach the foot of the tower in 3 seconds from the second observation.
(Alternatively, total time from the first observation is $\frac{d_1}{v}=9$ s; subtracting the elapsed 6 s also gives 3 s.)
Correct Answer: 3 seconds