1. The wind is blowing from the direction S40E (40 south of east) at a speed of 80 km/h. The pilot wants the ground speed to be 250 km/h at a bearing of N70E (70 east of north). Find the airspeed, use sine cosine law.

2. An airplane is flying with an airspeed of 400 km/h at a bearing of N20E. There is a wind blowing from the direction W30N (30 north of west) at a speed of 80 km/h. Inside the airplane, a passenger is walking at 5 km/h towards the front of the plane (relative to the plane). Determine the velocity of the passenger relative to the ground.

Solution: First, let's define the angles and directions: - Wind is from S40E, which is 40 south of east, or 50 east of south. - Desired ground speed is N70E, which is 70 east of north. The angle between the wind direction and the desired ground direction is: Angle = 180 50 70 = 60 Using the cosine law to find the airspeed V: V2 =V2+V2Z-2V,V, cos(60) Given: Ve =250km/h, V,, =80km/h Substituting values into the equation: 250% = V.2 + 80% 2V,, - 80 0.5 62500 = V2 4 6400 80V, V.2 80V, + 56100 = 0 Solving the quadratic equation: 804 /802 456100 o 2 80 6400 + 224400 sE= 80 /230800 - 9 _ 80+480.3 =g Va Ve Va Taking the positive root: V, = @ ~ 280.15km/h Thus, the airspeed is approximately 280.15 km /h. Problem 2: Determining Velocity of the Passenger Relative to the Ground Given: e Airspeed of the airplane: 400 km/h at N20E. Wind speed and direction: 80 km/h from W30N. e Passenger's walking speed inside the plane: 5 km/h towards the front of the plane. Solution: First, let's define the angles and directions: - Airspeed of the airplane is 400 km/h at N20E, which is 20 east of north. - Wind is from W30N, which is 30 north of west, or 150 from east. The angle between the airspeed direction and the wind direction is: Angle = 180 20 30 = 130 Using the cosine law to find the ground speed V: VE=VZ+V22V,V,cos(130) Given: . = 400km/h, V, = 80km/h Substituting values into the equation: V2 = 400% + 80 2 - 400 - 80 - cos(130) Vy = 160000 + 6400 + 2 - 400 - 80 - 0.6428 V2 = 160000 + 6400 + 51424 V}? =217824 Vg = V217824 ~ 466.58 km /h Now, let's add the passenger's speed relative to the plane: V, = 5km/h Since the passenger is walking towards the front of the plane, the total speed relative to the ground Vg = Vg4V, = 466.58 4 5 = 471.58 km/h Thus, the velocity of the passenger relative to the ground is approximately 471.58 km/h