Answer:
[tex]301.29\ \text{mph}[/tex]
[tex]119.5^{\circ}[/tex]
Step-by-step explanation:
[tex]v_p[/tex] = Velocity of plane = 320 mph
[tex]v_w[/tex] = Velocity of wind = 35 mph
[tex]\theta[/tex] = Angle between wind and plane directions = [tex]125^{\circ}[/tex]
From triangle law we have resultant
[tex]v_r=\sqrt{v_p^2+v_w^2+2v_pv_w\cos\theta}\\\Rightarrow v_r=\sqrt{320^2+35^2+2\times 320\times 35\cos125^{\circ}}\\\Rightarrow v_r=301.29\ \text{mph}[/tex]
Direction is given by
[tex]\phi=\tan^{-1}\dfrac{v_p\sin\theta}{v_w+v_p\cos\theta}\\\Rightarrow \phi=\tan^{-1}\dfrac{320\sin125^{\circ}}{35+320\cos125^{\circ}}\\\Rightarrow \phi=-60.46^{\circ}=180-60.46=119.5^{\circ}[/tex]
The magnitude of the plane is [tex]301.29\ \text{mph}[/tex] moving at an angle of [tex]119.5^{\circ}[/tex].
