# Finding the length

• May 30th 2009, 07:44 PM
Judi
Finding the length
Radar stations A and B are on an east-west line, 8.6 km apart. Station A detects a plane at C, on a bearing of 53 degree. Station B simultaneously detects the same plane, on a bearing of 323 degree. Find the distance from B to C.

That is the question. And Why can I not use cosine to find the lenth?

And I thank you in advance everyone who help me with this problem.
• May 30th 2009, 08:07 PM
Soroban
Hello, Judi!

What do you mean by "use cosine"?

Quote:

Radar stations $A$ and $B$ are on an east-west line, 8.6 km apart.
Station $A$ detects a plane at $C$, on a bearing of 53°.
Station $B$ simultaneously detects the same plane, on a bearing of 323°.
Find the distance from $B$ to $C.$

Why can I not use cosine to find the length?

Code:

                        C                         *       :              *  *    :       :          *      *37°:       :        *          *  :       : 53° *              * :       :  * 37°          53° *:       * - - - - - - - - - - - *       A          8.6          B

If you mean "the Law of Cosines", we can't.
. . The Law of Cosines requires two sides ... and we have only one.

If you note that $\angle C = 90^o$, you can use cosine.

. $\cos B \:=\:\frac{BC}{AB} \quad\Rightarrow\quad BC \:=\:8.6\cos53^o \quad\hdots\:\text{etc.}$

• May 31st 2009, 02:48 PM
bjhopper
Station bearing on plane posted by Judi
the east and west station should be identifed by a letter and station. I assumed that A was east and Bwest.I do not get C being 90 deg.using the bearing at A 323 and the bearing at B 53 .the relative bearing at A is 63 deg north of west. This makes Abc an isosoles triangle.BC is obtained using the cosine function.

bjh
• May 31st 2009, 04:44 PM
aidan
Quote:

Originally Posted by bjhopper
the east and west station should be identifed by a letter and station. I assumed that A was east and Bwest.I do not get C being 90 deg.using the bearing at A 323 and the bearing at B 53 .the relative bearing at A is 63 deg north of west. This makes Abc an isosoles triangle.BC is obtained using the cosine function.

bjh

If station A were east of station B then the bearing angles are out of phase by 180 degrees.

Soroban's beautiful image depicts the facts.

$\angle CAB = 37^o$
$\angle ABC = 53^o$
that leaves
$\angle BCA = 180^o - 37^o -53^o = 90^o$

• May 31st 2009, 05:21 PM
Judi
Thank you for your drawing. Yes, I meant by the method 2 you showed. I know I can get the answer by using either cos or sin. But is it work every single time? I have encountered the problem where cos didn't work.

Quote:

Originally Posted by Soroban
Hello, Judi!

What do you mean by "use cosine"?

Code:

                        C                         *       :              *  *    :       :          *      *37°:       :        *          *  :       : 53° *              * :       :  * 37°          53° *:       * - - - - - - - - - - - *       A          8.6          B

If you mean "the Law of Cosines", we can't.
. . The Law of Cosines requires two sides ... and we have only one.

If you note that $\angle C = 90^o$, you can use cosine.

. $\cos B \:=\:\frac{BC}{AB} \quad\Rightarrow\quad BC \:=\:8.6\cos53^o \quad\hdots\:\text{etc.}$