Question

an airplane is flying from montreal to vancouver. the wind is blowing from the west at 60km/h. the airplane flies at an airspeed of 750km/h and must stay on a heading of 65 degrees west of north. what heading should the pilot take to compensate for the wind? what is the speed of the airplane relative to the ground?

Answer 1

Refer to the diagram shown below.

In order for the airplane to maintain a heading of 65° west of north,  it actually heads at (65° +x) west of north at a speed of 750 km/h, to compensate for the wind blowing from west t east at 60 km/h.

The actual speed of the airplane relative to the ground is V km/h.

From geometry, obtain the triangle shown.
The speed of 750 km/h is opposite an angle of 155°, and the unknown angle x is opposite the wind speed of 60 km/h.

From the Law of Sines, obtain
$\frac{sin(x)}{60}= \frac{sin(155^{o}}{750} \\ sin(x)= (\frac{60}{750})sin(155^{o})=0.0338$
$x=sin^{-1}0.0338=1.937^{o}$

The heading that the pilot should take is 65 + 1.937 = 66.94° west of north.

The third angle of the triangle is 180 - (155 + 1.937) = 23.063°.
Use the Law of Sines to calculate V.
$\frac{V}{sin(23.063^{o})}= \frac{750}{sin(155^{o})}$
$V=( \frac{sin(23.063)}{sin(155)} )750=695.21\,km/h$

Answer:
The pilot heads approximately 67° (nearest integer) west of north.
The speed of the airplane relative to ground is 695.2 km/h (nearest tenth)