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
[tex] \frac{sin(x)}{60}= \frac{sin(155^{o}}{750} \\ sin(x)= (\frac{60}{750})sin(155^{o})=0.0338[/tex]
[tex]x=sin^{-1}0.0338=1.937^{o}[/tex]
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.
[tex] \frac{V}{sin(23.063^{o})}= \frac{750}{sin(155^{o})} [/tex]
[tex]V=( \frac{sin(23.063)}{sin(155)} )750=695.21\,km/h[/tex]
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)
