Question

A bird was flying in a line parallel to the ground from north to south at a height of 200 metres. Tom, standing in the middle of the field, first he observed the bird in the north at an angle of 30^o. After 3 minutes, he again observed it in the south at an angle of 45^o. Find the speed of the bird in kilometres per hour.

Answer 1

Let the bird initially be at point B and let it reach point C after 3 minutes.
Also, let Tom stand at point A.
Thus, we have:

$$\begin{gathered} \mathrm{In}△ADB, \\ \tan 30°=\frac{AD}{BD} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{200}{BD} \\ \Rightarrow BD=200\sqrt{3}m...\left(\mathrm{i}\right) \\ \mathrm{In}△ACD, \\ \tan 45°=\frac{AD}{DC} \\ \Rightarrow 1=\frac{200}{DC} \\ \Rightarrow DC=200\mathrm{m}...\left(\mathrm{ii}\right) \\ \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ BC=BD+DC=200\sqrt{3}+200\mathrm{m} \\ =546m=0.546\mathrm{km} \\ \mathrm{The}\mathrm{distance}\mathrm{BC}\mathrm{is}\mathrm{covered}\mathrm{in}3\mathrm{minutes}. \\ \therefore BC=3\times \mathrm{Speed}\mathrm{of}\mathrm{the}\mathrm{bird} \\ \mathrm{Or}, \\ 0.546=\frac{3}{60}\times \mathrm{Speed} \\ \mathrm{Or}, \\ \mathrm{Speed}=10.92\mathrm{km}/\mathrm{h} \end{gathered}$$