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

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)