A boat which has a speed of 5kmh^ - 1 in still water crosses a...

$t = \dfrac{15}{60} = \dfrac{1}{4}h$

$∴ {V}_{b}cos\theta = \dfrac{1}{( \dfrac{1}{4} )} = 4kmh^{- 1}$

$$\begin{array}{ll}⇒ cos\theta = \dfrac{4}{{V}_{b}} = \dfrac{4}{5} \\ ⇒ sin\theta = \dfrac{3}{5}\end{array}$$

By velocity triangle ABC, $sin\theta = \dfrac{{V}_{r}}{{V}_{b}}$

$$\begin{array}{ll}∴ \dfrac{Vr}{Vb} = \dfrac{3}{5} \\ ⇒ \dfrac{Vr}{5} = \dfrac{3}{5} \\ ⇒ {V}_{r} = 3kmh^{- 1}\end{array}$$