# Thread: Reaaly stuck on this trig problem!

1. ## Reaaly stuck on this trig problem!

Two cities A and B are 300 miles apart. In flying from A to B, a pilot inadvertently took a course that was 15 degrees in error. The error was discovered after 15 minutes at a constant speed of 400 mph

a. how far is the plane from B?
b. through what angle should the pilot turn to correct the course?
c.what new constant speed should be maintained so that no time is lost due to the error?

A picture is given Triangle ABC is given where side AB= 300 and angle A =15. Any help please!

2. I would use coordinate geometry here. If we locate city A at the origin, and put city B at (300,0), and let point C be the point at which the pilot discovers his/her error, what are the coordinates of C?

3. Hello, tballer!

Two cities A and B are 300 miles apart.
In flying from A to B, a pilot inadvertently took a course that was 15 degrees in error.
The error was discovered after 15 minutes at a constant speed of 400 mph.
Code:
                              . D
C   .
o
100   *     *  x
*           *
* 15o             *
A o-----------------------o B
: - - - -  300  - - - - :

The plane intended to fly directly east from A to B at 400 mph.
The flight should have taken $\frac{300}{400} = \frac{3}{4}$ hour.
Instead it flew 15o north of east for 15 minutes to point C.
. . $\angle A \,=\,15^o,\;AC = 100$ miles.
Let $x = CB.$

a. How far is the plane from B?

Law of Cosines: .$x^2 \;=\;100^2 + 300^2 - 2(100)(300)\cos15^o \;=\;44,\!144.45042$

Therefore: .$x \;=\;210.1058077 \;\approx\;210.1\text{ miles.}$

b. Through what angle should the pilot turn to correct the course?

Let $C = \angle ACB.$

Law of Cosines: .$\cos C \;=\;\dfrac{100^2 + 210.1^2 - 300^2}{2(100)(210.1)} \;=\;-0.853355307$

Hence: .$C \:=\:148.5676649^o \:\approx\:148.6^o$

Therefore: .$\angle DCB \:=\:180^o - 148.6^o \;=\;31.4^o$

The pilot should turn 31.4o to the south.

c. What new constant speed should be maintained so that no time is lost due to the error?

The pilot has a half-hour left to fly the 210.1 miles to B.

His speed must be: $\dfrac{210.1}{0.5} \:=\:420.2\text{ mph.}$