An aircraft takes off from an airstrip at an average speed of 20km/hr on a bearing of 052° for three hours it then changes course and flies on a bearing of 028° at an average speed of 30km/he for another 1½ hour Find it's distance from the starting point The bearing of the aircraft from the airstrip

The aircraft flies at 20 km/hr for 3 hours on a bearing of 052°.
Distance (d₁) = Speed × Time
d₁ = 20 km/hr × 3 hr = 60 km

North component (y₁) = d₁ × cos(Bearing)
y₁ = 60 km × cos(52°)
y₁ ≈ 60 km × 0.6157 = 36.942 km

East component (x₁) = d₁ × sin(Bearing)
x₁ = 60 km × sin(52°)
x₁ ≈ 60 km × 0.7880 = 47.280 km

The aircraft flies at 30 km/hr for 1.5 hours on a bearing of 028°.
Distance (d₂) = Speed × Time
d₂ = 30 km/hr × 1.5 hr = 45 km

North component (y₂) = d₂ × cos(Bearing)
y₂ = 45 km × cos(28°)
y₂ ≈ 45 km × 0.8829 = 39.7305 km

East component (x₂) = d₂ × sin(Bearing)
x₂ = 45 km × sin(28°)
x₂ ≈ 45 km × 0.4695 = 21.1275 km

Total North displacement (Y) = y₁ + y₂
Y = 36.942 km + 39.7305 km = 76.6725 km

Total East displacement (X) = x₁ + x₂
X = 47.280 km + 21.1275 km = 68.4075 km

Distance = √(X² + Y²)
Distance = √((68.4075 km)² + (76.6725 km)²)
Distance = √(4679.58 km² + 5878.68 km²)
Distance = √(10558.26 km²)
Distance ≈ 102.75 km

2) The bearing of the aircraft from the airstrip

Since both X (East) and Y (North) components are positive, the aircraft is in the North-East quadrant. The bearing is the angle measured clockwise from North.
Angle (θ) = arctan(East component / North component)
θ = arctan(X / Y)
θ = arctan(68.4075 / 76.6725)
θ = arctan(0.8921)
θ ≈ 41.72°

The bearing is written with three digits.
Bearing = 041.7°