Hello, Jim!

I hope you understand "bearings" . . .

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

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

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

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