Question

By using degree measurements to represent compass directions,
you can describe the heading, or direction, in which a plane is
traveling. In this system, 0° (360°) corresponds to due north,
90° corresponds to due east, 180° corresponds to due south,
and 270° corresponds to due west.
Angles are measured in a clockwise direction. This is different
from measuring angles in standard position on a coordinate
system.
With this method you can describe a flight path in terms of
distances and headings. For example, suppose a plane flies 300 mi
on a heading of 45°; then the plane changes course and flies 200 mi
on a heading of 150°.
If you could determine the angle between the two “legs” of
the trip (B), you could then use the Law of Cosines to find
how far the plane has traveled from its point of departure (b).
Because is parallel to , B is supplementary to BAD, which
is 105°. Thus B = 180° – 105° = 75°. Using the law of cosines,
b2 = 3002 + 2002 – 2(300)(200) cos 75°, and b = 314.6 mi.

Determine the heading on which the plane would have to travel to return to point A.

1. First use the Law of Sines to find BAC, and then add 45°.


2. Add 180°, to find the return heading.



Suppose a plane flies 240 mi on a heading of 35°, Then the plane changes course and flies 160 mi on a heading of 160°.
3. Determine the heading on which the plane would have to travel to return to its point of origin.

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:

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

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