# Math Help - Finding the length

1. ## 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.

2. Hello, Judi!

What do you mean by "use cosine"?

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.

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

3. ## 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

4. 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
$\angle BCA = 180^o - 37^o -53^o = 90^o$

5. 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.

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.

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