A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?

Answer: Option B

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, ∠ ADC = 30° , ∠ ACB = 45°
Let AB = h, BC = x, CD = y
$$\begin{array}{rl}& \tan 45^{\circ }=\frac{AB}{BC}=\frac{h}{x}\\ & \Rightarrow 1=\frac{h}{x}\\ & \Rightarrow h=x......\left(1\right)\\ & \tan 30^{\circ }=\frac{AB}{BD}\\ & =\frac{AB}{\left(BC+CD\right)}\\ & =\frac{h}{x+y}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+y}\\ & \Rightarrow x+y=\sqrt{3}h\\ & \Rightarrow y=\sqrt{3}h-x\end{array}$$
$$\Rightarrow y=\sqrt{3}h-h$$     (∵ Substituted the value of x from equation 1 )
$$\Rightarrow y=h\left(\sqrt{3}-1\right)$$
Given that distance y is covered in 8 minutes.
i.e, distance $$h\left(\sqrt{3}-1\right)$$   is covered in 8 minutes.
Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).
Let distance h is covered in t minutes.
since distance is proportional to the time when the speed is constant, we have
$$\begin{array}{rl}& h\left(\sqrt{3}-1\right)\propto 8......\left(\text{A}\right)\\ & h\propto t..............\left(\text{B}\right)\\ & \frac{\left(\text{A}\right)}{\left(\text{B}\right)}\Rightarrow \frac{h\left(\sqrt{3}-1\right)}{h}=\frac{8}{t}\\ & \Rightarrow \left(\sqrt{3}-1\right)=\frac{8}{t}\\ & \Rightarrow t=\frac{8}{\left(\sqrt{3}-1\right)}\\ & =\frac{8}{\left(1.73-1\right)}\\ & =\frac{8}{.73}\\ & =\frac{800}{73}\text{minutes}\\ & =10\frac{70}{73}\text{minutes}\\ & \approx \text{10}\text{minutes}\text{57}\text{seconds}\end{array}$$