Can you show the workings for the last part (that you explained saying you just use the distance formula and then you have your answer which is 389.82)?

Answer 1

If the wind looks like this:

then the portion of the wind traveling in the x-direction or east (shown in my diagram as V_W-E) could be calculated using cos(30°) = adjacent/hypotenuse = V_W-E/100 or 100cos(30°) = V_W-E

Using your calculator, V_W-E = 86.6 mph

The portion of the wind traveling in the y-direction or north (shown in my diagram as V_W-N) could be calculated using sin(30°) = opposite/hypotenuse = V_W-N/100 or 100sin(30°) = V_W-N

Using your calculator, V_W-N = 50 mph.

These add to the velocity of the airplane which is just 300 going east. The wind in the east direction just adds directly so the new velocity of the airplane going east is 300 + 86.6 = 386.6 mph. But the airplane now has a component of velocity in the north direction from the wind of 50 mph. To find the resultant, you need to consider these lengths as two legs of a right triangle then the new velocity is the hypotenuse.

v² = 386.6² + 50² = 151961

v = √151961 = 389.82 mph although I'd round it off to at least 390 mph

Answer 2

Assuming the 30° is relative to East.

Draw and label a diagram

The angle between the plane and the wind is 60°

Use Law of Cosines

s² = 300² + 100² - 2(300)(100) cos 60°