a pilot must fly a destination 520 km away in a direction of N30 degree E. The air speed of the plane is 320Km and the wind speed is 85 km/h E20 degree Suth. Find the heading of the plane and the time for the trop.

Answer 1

This question assumes knowledge of vectors and how to solve for a triangle using the sine rule. If these subjects are vague, a revision is in order.

Let the plane start at O (origin) in the x-y plane, x=East, and y=North. Use a piece of graphics paper to help you draw the sketch according to the following directions. Draw it roughly to scale, but it does not have to be accurate, as we will solve the unknowns by trigonometry.

We will consider a time period of 1 hour, hence the plane travels 320 km and the wind, 85 km. We take the vectorial sum of the effect of wind and the plane, the direction of wihci is yet unknown.

Draw a line from O in the direction of the destination, namely N30Em, length about 400 units (km).

Draw the wind vector OA, in the direction E20S and a magnitude of 85 km/h.

From A, draw a vector AB, of magnitude 320 and intersecting the line in the N30E direction at B.

If we isolate the triangle OAB, we have:
Angle BOA = (90-30)+20 = 80°
OA = 85 km

Using sine rule,
sin(B)/85 = sin(80)/320
solve to get B=15.1644°

Since angles O + A + B =180,
we solve for A=180-O-B = 84.8356&deg.

Using sine rule again,
OB/sin(A) = 320/sin(80)
we solve for OB=324 km/hr approx.

Heading: NθE where θ=30-∠B=15° approx.

Answer 2

From A, draw a vector AB, of magnitude 320 and intersecting the line between O and destination.

If we isolate the triangle OAB, we have:
Angle BOA = (90-30)+20 = 80°
OA = 85 km
OB = 320 km