A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?

Answer: Option A

Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, ∠ ACB = 45°, ∠ ADC = 30°, BC = 100 m
$$\begin{array}{rl}& \tan 45^{\circ }=\frac{AB}{BC}\\ & \Rightarrow 1=\frac{AB}{100}\\ & \Rightarrow AB=100......\left(eq:1\right)\end{array}$$
$$\tan 30^{\circ }=\frac{AB}{BD}$$
$$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{BD}$$     (∵ Substituted the value of AB from equation 1)
$$\Rightarrow BD=100\sqrt{3}$$
$$\begin{array}{rl}& CD=\left(BD-BC\right)\\ & =\left(100\sqrt{3}-100\right)\\ & =100\left(\sqrt{3}-1\right)\end{array}$$
It is given that the distance CD is covered in 10 seconds.
i.e., the distance $$100\left(\sqrt{3}-1\right)$$     is covered in 10 seconds.
$$\begin{array}{rl}& \text{Required}\text{speed}\\ & =\frac{\text{Distance}}{\text{Time}}\\ & =\frac{100\left(\sqrt{3}-1\right)}{10}\\ & =10\left(1.73-1\right)\\ & =7.3\text{meter/seconds}\\ & =7.3\times \frac{18}{5}\text{km/hr}\\ & =26.28\text{km/hr}\end{array}$$