An aircraft flies at $400 \mathrm{km}/\mathrm{h}$ in still air. A wind of $200\sqrt{2} \mathrm{km}/\mathrm{h}$ is blowing from the south towards north. The pilot wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if $AB=1000 \mathrm{km}$.

Given that $v_w= 200\sqrt{2} \mathrm{km}/\mathrm{h}, {\mathrm{v}}_{\mathrm{aw}}=400 \mathrm{km}/\mathrm{h}$ and $v_a$ should be along AB or in north-east direction. 

Thus, the direction of $v_{aw}$ should be such as resultant of $v_w$ and $v_{aw}$ is along $AB$ or in north-east direction. 

Let $v_a$ makes as angle $\alpha$ with $AB$ as shown in the above figure. 

Applying sine law in triangle $ABC$, we get 

$$\begin{gathered} \frac{CB}{\sin 45°}=\frac{AC}{\sin \alpha } \\ \sin \alpha =\left(\frac{AC}{CB}\right)\sin 45° \\ =\left(\frac{200\sqrt{2}}{400}\right)\frac{1}{\sqrt{2}}=\frac{1}{2} \\ \alpha =30° \end{gathered}$$

Therefore, the pilot should steer in a direction at an angle of $\left(45°+\alpha \right)\text{ or }75°$ from north  towards east. 

Further, $$\begin{gathered} \frac{\left|v_a\right|}{\sin \left(180°-45°-30°\right)}=\frac{400}{\sin 45°} \\ \left|v_a\right|=\frac{\sin 105°}{\sin 45°}\times \left(400\right)\mathrm{km}/\mathrm{h} \\ =\left(\frac{0.9659}{0.707}\right)\left(400\right)\mathrm{km}/\mathrm{h} = 546.47 \mathrm{km}/\mathrm{h} \end{gathered}$$

Therefore, the time of journey from A to B is 

$t=\frac{AB}{\left|{\mathbf{v}}_a\right|}=\frac{1000}{546.47} \mathrm{h}=1.83 \mathrm{h}$