# Thread: Another vector application problem

1. ## Another vector application problem

Hello, somewhere along the line I got stuck again. Again, any help would be really appreciated! This time the problem didn't state much that's why I'm in a dilemma.

"An air traffic patrol helicopter hovering at C sights a vehicle at A. One minute later the vehicle is at B. points A, B and C are int he same vertical plane. Find the speed of the vehicle in miles per hour."

Here's the picture that follows it.

PS. If you are befuddle by the picture, the 72° is from the top horizontal line to line CA while the 37° is from top horizontal line to line CB. I tried to make it look as neat as possible so sorry if you think it is messy.

2. Hello, wanderlust!

An air traffic patrol helicopter hovering at $\displaystyle C$ sights a vehicle at $\displaystyle A.$
One minute later the vehicle is at $\displaystyle B.$
Find the speed of the vehicle in miles per hour.
Code:
- - - - - - - - - - - o C
37°  * *|
*35°* |
*     *  |
*       *18°|
*         *    | 3000
*           *     |
*             *      |
*               *       |
* 37°        108° * 72°    |
o - - - - - - - - - o - - - - *
B         x         A         D

Let: $\displaystyle x = AB$
We can determine all the angle in the problem.
. .
(I hope you followed that.)

In right triangle $\displaystyle CDA\!:\;\;\sin72^o \,=\,\frac{3000}{AC} \quad\Rightarrow\quad AC \:=\:\frac{3000}{\sin72^o} \:\approx\:3154.4$

In $\displaystyle \Delta CAB\text{, the Law of Sines: }\;\frac{x}{\sin35^o} \:=\:\frac{AC}{\sin37^o} \quad\Rightarrow\quad x \:=\:\frac{3154.4\,\sin35^o}{\sin37^o} \:\approx\:3006.4$

Hence, the card drove 3006.4 feet in one minute.

That's 180,384 feet in one hour,
. . or about 34.2 miles in one hour.

The speed of the car is 34.2 mph.