Given ABC = 50^ and ACB = 60^.

Here's the solution for the bearing problem: a) Calculate the bearing of A from B Step 1: Find the angle BAC in ABC. The sum of angles in a triangle is 180^. Given ABC = 50^ and ACB = 60^. BAC = 180^ - ( ABC + ACB) BAC = 180^ - (50^ + 60^) BAC = 180^ - 110^ = 70^ Step 2: Determine the bearing of C from B. Since A is due north of C, the line segment CA is a North-South line. Draw a North line at B, parallel to CA. Let N_B be the North direction from B. The angle ACB = 60^ and the angle between the South line from B (let's call it S_B) and BC are alternate interior angles. So, the angle from the South line at B to BC is 60^. The bearing of C from B is measured clockwise from the North line at B. This is 180^ (to the South line) plus the 60^ angle. Bearing of C from B = 180^ + 60^ = 240^ Step 3: Calculate the bearing of A from B. The bearing of A from B is the bearing of C from B minus the angle ABC, because A is 50^ counter-clockwise from C, relative to B. Bearing of A from B = Bearing of C from B - ABC Bearing of A from B = 240^ - 50^ = 190^ The bearing of A from B is 190^. b) Calculate the bearing of B from A Step 1: Identify the North line at A. Since A is due north of C, the line segment AC is a North-South line. The North direction from A points away from C. The line segment AC itself represents the South direction from A. Step 2: Calculate the bearing of B from A. The bearing of B from A is measured clockwise from the North line at A to the line AB. The angle from the North line at A, clockwise to the South line at A (which is AC), is 180^. We found BAC = 70^. This angle is between the South line (AC) and the line AB. Since AB is 70^ to the "left" (west) of the South line (AC), we subtract this angle from 180^. Bearing of B from A = 180^ - BAC Bearing of B from A = 180^ - 70^ = 110^ The bearing of B from A is 110^. 3 done, 2 left today. You're making progress.