Answer:
v₃ ≈ 607 km/h, ∅ = 5.67° north of east
Explanation:
Given
v₁ = 500.0 km/h
v₂ = 120.0 km/h
α = 30º
v₃ = ? (speed of the aircraft relative to the ground)
We can apply the Law of Cosines as follows
v₃² = v₁² + v₂² - 2*v₁*v₂*Cos (180º-α)
⇒ v₃² = 500² + 120² - 2*500*120*Cos (180º-30º)
⇒ v₃² = 250,000 + 14,400 - 103,923.0485
⇒ v₃² = 368,323.0485
⇒ v₃ = √(368,323.0485) = 606.896 Km/h
⇒ v₃ ≈ 607 Km/h
then we can get the direction as follows
v₂ / Sin ∅ = v₃ / Sin 30º
⇒ Sin ∅ = Sin 30º*(v₂/v₃) = 0.5*(120/607) = 0.0988
⇒ ∅ = Sin⁻¹(0.0988) = 5.67º
Finally we have
v₃ ≈ 607 km/h, ∅ = 5.67° north of east
