Question

An athlete runs at a velocity of 18m\s due East. A strong wind travelling at 8m\s blows 230° against him. Find the resultant vector using triangle of vectors

Answer 1

The law of cosines gives

$$v^2=18^2+8^2-2\times 18\times 8\times \cos 50^{\circ}=202.877$$

$$v=14.2\:\rm{m/s}$$

The law of sines gives

$$\frac{\sin\theta}{8}=\frac{\sin 50^{\circ}}{14.2}$$

$$\sin\theta =\frac{\sin 50^{\circ}}{14.2}\times 8=0.432$$

$$\theta=\sin^{-1}(0.432)=25.6^{\circ}\quad (\rm{South\: of\: East})$$