Ajet moving at 500.0 km/h due east is in a region where the wind is moving at 120.0 km/hr in a direction 30.0 degrees north of east. what is the speed of the aircraft relative to the ground?

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

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