The diagram above shows 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 is 026 degrees and the bearing of C from B is 328 degrees. Calculate the distances AC and BC
The diagram above shows 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 is 026 degrees and the bearing of C from B is 328 degrees. Calculate the distances AC and BC
Let's get some angles first. Angle A is the complement of 26 degrees, so it's equal to 64 degrees. Angle B is the complement of what's left over from 360 - 328 degrees. (Sorry I can't think of a better description of it.) So it's the complement of 360 - 328 = 32 degrees, ie. angle B is 58 degrees. We know two angles of the triangle, so now we can find the third. Angle C will be 180 - A - B = 180 - 64 - 58 = 58 degrees.
So this triangle is isosceles. Thus AC = AB = 50 m. We can use either the Law of Sines or Law of Cosines to find BC. I'll use the Law of Sines:
$\displaystyle \frac{AB}{sin(C)} = \frac{BC}{sin(A)}$
$\displaystyle \frac{50}{sin(58)} = \frac{BC}{sin(64)}$
$\displaystyle BC = \frac{50}{sin(58)} \cdot sin(64) = 52.9919 \, m$
I'll leave it to you to verify that the Law of Cosines gives the same answer.
-Dan