## Trigonometry (11th Edition) Clone

The ground speed is 380.4 mph The actual bearing of the plane is $64^{\circ}$
Let $a = 355~mph$ Let $b = 28.5~mph$ Let angle $\theta$ be the angle between these two vectors. Then $\theta = 90^{\circ}-62^{\circ} = 28^{\circ}$ Let $c$ be the resultant of these two vectors. Note that the magnitude of $c$ is the ground speed. We can use the parallelogram rule to find $c$: $c = \sqrt{a^2+b^2+2ab~cos~\theta}$ $c = \sqrt{(355~mph)^2+(28.5~mph)^2+(2)(355~mph)(28.5~mph)~cos~28^{\circ}}$ $c = \sqrt{144703.7~mph^2}$ $c = 380.4~mph$ The ground speed is 380.4 mph We can use the law of sines to find the angle B between the ground speed vector and the airspeed vector: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~C}{c}$ $B = arcsin(\frac{b~sin~C}{c})$ $B = arcsin(\frac{(28.5)~sin~152^{\circ}}{380.4})$ $B = arcsin(0.03517)$ $B = 2.0^{\circ}$ The actual bearing of the plane is $62^{\circ}+2.0^{\circ} = 64^{\circ}$