# find distance

• December 10th 2006, 02:05 AM
jim49990
find distance
i have a triangular plot of land through which a stream runs. the point B is due east of A and AB is 50m long. the bearing of C from A 026 degrees and the bearing of C from B is 328 degrees. calculate the distances AC and BC.

could this problem be solved in a alternative way? if so, describe an alternative method.
thanks
jim
• December 10th 2006, 03:13 AM
Fredrik
Quote:

Originally Posted by jim49990
i have a triangular plot of land through which a stream runs. the point B is due east of A and AB is 50m long. the bearing of C from A 026 degrees and the bearing of C from B is 328 degrees. calculate the distances AC and BC.

could this problem be solved in a alternative way? if so, describe an alternative method.
thanks
jim

If "bearing" means what I think it means, the interior angle of the triangle at A is 26 degrees and the interior angle of the triangle at B is 32 degrees. Is that right? Edit: No! The angles are 64 degrees and 58 degrees. In that case, I suggest you draw a line from C to a point D on the AB line chosen so that AB and CD are perpendicular.

CD/DB=tan(32 degrees)

Solve this for CD and use that CD=BC sin(32 degrees) and CD=AC sin(26 degrees).

Alternative method? This is the only one I can think of right now.
• December 10th 2006, 04:13 AM
jim49990
i dont think that is right
i think the answer is 50m
• December 10th 2006, 05:15 AM
Soroban
Hello, Jim!

I hope you understand "bearings" . . .

Quote:

I have a triangular plot of land through which a stream runs.
Point B is due east of A and AB is 50m long.
The bearing of C from A is 26° and the bearing of C from B is 328°.
Calculate the distances AC and BC.

Code:

                        C                         *                     *  *       :          *  58°  *  :       :        *          *  :       : 26° *              * :       :  * 64°          58° *:     A * - - - - - - - - - - - * B                   50

Since $\angle A = 64^o$ and $\angle B = 58^o$, then: $\angle C = 58^o$

. . $\Delta ABC$ is isosceles: $AC = AB = 50$ m.

Law of Cosines: . $BC^2\:=\:50^2 + 50^2 - 2(50)(50)\cos64^o\:=\:2808.144266$

Therefore: . $BC \:=\:52.99192642\:\approx\:53$ m.

• December 10th 2006, 05:23 AM
Fredrik
Apparently I got the angles of the triangle wrong. The method I suggested will get you the correct answer if you use the correct angles: 64 degrees and 58 degrees. So you can consider that the "alternative" method.
• December 10th 2006, 05:27 AM
jim49990
many thanks
thanks for that. the fact that there is a stream is irrelavant..
jim