Cycling speed = 8 m/s \ \

Cycling direction = north \ \

Wind blowing speed =  3 m/s \ \

Wind direction = south-east \ \

The picture representation of problem is \ \

\ \

Consider \ \

a = OA = CB = Cycling speed = 8 m/s \ \

b = OB = AC = Wind speed = 3 m/s \ \

c = OC = relative speed \ \

Angle C = ∠AOC = ∠AOE + ∠BOE \ \

Angle C = 90 + 45 = 135^o \ \

So the relative velocity can be calculated using cosine rule. \ \

c² = 8² + 3² -2(8)(3) cos(45°)  \ \

c² = 64 +9 - 47.81 \ \

c² = 30.05 \ \

c = 6.25 m/s \ \

The relative velocity  is 6.25 m/s . \ \

The direction of two cars can be calculated using formula : . \ \

cos(∠AOC) = 0.94 \ \

∠AOC = 19.8^o \ \

the resultant velocity of the cyclist in direction is 16.58° right side to north side. \ \

Solution : \ \

The relative velocity  is 6.25 m/s . \ \

the resultant velocity of the cyclist in direction is 16.58° right side to north side.