Question

By using degree measurements to represent compass directions,

Answer 1

Please find the attached diagram which best represents the information given in the question.

From the diagram it is clear that after taking the turn and having a heading of $160^0$, the plane makes an angle $\angle PBC=125^0$ as shown in the diagram. This, obviously, makes $\angleABC=55^0$ by making use of the fact that $\angle PBC$ and $\angle ABC$ are supplementary.

Now, using the Cosine Formula as shown in the question example we can find AC to be:

$AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)Cos(m\angle ABC)}$

Thus, $(AC)=\sqrt{(240)^2+(160)^2-2(240)(160)Cos(55^0)}$

$\therefore AC\approx 197.86$ miles

Now, using the Sine Formula for a triangle, we can find the angle $m\angle BAC$ as:

$\frac{Sin(m\angle BAC)}{BC} =\frac{Sin(m\angle ABC)}{AC}$

$\frac{Sin(m\angle BAC)}{160} =\frac{Sin(55^0)}{197.86}$

$\therefore Sin(m\angle BAC)=\frac{160}{197.86}\times Sin(55^0)$

Thus, $(m\angle BAC)=Sin^{-1}(\frac{160}{197.86}\times Sin(55^0))\approx 41.48^0$

Thus, all that we have to do to find the return heading of the plane is to add $35^0$ to $41.48^0$ and then we will add $180^0$ to it.

Thus, the plane's return heading is:

$35^0+41.48^0+180^0\approx 256.48^0$

Part 1

We know that $\angle ABC=75^0$ and AC=314.6 miles.

Therefore, we get:

$\frac{Sin(75^0)}{314.6} =\frac{Sin(m\angle BAC)}{200}$

$\therefore \angle BAC)=Sin^{-1}(\frac{200}{314.6}\times Sin(75^0))\approx37.88^0$