A passenger flies on a heading of N40W with an airspeed of 240 km/h. His actual ground velocity is 250 km/h at a bearing of N60E. the speed and direction of the wind are V=245.15kw/h at N83.46E. This is further explained below.

<h3>What is the speed and direction of the wind.?</h3>

Generally, the equation for the cosine rule is mathematically given as

a2 = b2 + c2 – 2bc cos ∠x.

Therefore

V^2 = (240^2 + 250^2) – (2(240*250)* cos 100)

V^2=140937.7813

V=375.42kw/h

In conclusion, the sine rule is mathematically given as

Sin x/240=sin60/375.42kw/h

Therefore

sin x=sin100*250/375.42kw/h

x=40.981

Hence

$\theta= 180-2*48.27$

Direction N83.46E