# Another vector application problem

• February 22nd 2010, 07:44 PM
wanderlust
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.
http://i46.tinypic.com/2mwa2jo.jpg
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.
• February 22nd 2010, 10:23 PM
Soroban
Hello, wanderlust!

Quote:

An air traffic patrol helicopter hovering at $C$ sights a vehicle at $A.$
One minute later the vehicle is at $B.$
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: $x = AB$
We can determine all the angle in the problem.
. .
(I hope you followed that.)

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

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

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.