1. Statement of the problem. Problem: Points A, B, C lie on level ground with C due east of B, $\angle ABC=80^\circ$ and $\angle ACB=20^\circ$. Calculate the bearing of (a) C from A and (b) B from A. 2. Find the remaining angle. Using angle sum in a triangle, $\angle A=180^\circ-80^\circ-20^\circ=80^\circ$. 3. Choose coordinates and scale. Place B at $(0,0)$ and C at $(1,0)$ so that $BC=1$; this choice fixes scale and preserves bearings. 4. Find side lengths by the Law of Sines. With $BC=1$ opposite $\angle A=80^\circ$, the Law of Sines gives $AB=\dfrac{\sin20^\circ}{\sin80^\circ}\approx0.3472963553$. 5. Find coordinates of A. From B the side BA has length $AB$ and makes an angle $80^\circ$ above east, so $A=(AB\cos80^\circ,AB\sin80^\circ)$. Numerically $A\approx(0.060307,0.342020)$. 6. Vector from A to C and its components. $\vec{AC}=C-A=(1-0.060307,0-0.342020)\approx(0.939693,-0.342020)$. 7. Bearing of C from A. Bearing is measured clockwise from north and can be computed by $\theta=\operatorname{atan2}(\Delta x,\Delta y)$ where $\Delta x$ is east and $\Delta y$ is north. Here $\theta=\operatorname{atan2}(0.939693,-0.342020)\approx110^\circ$. Thus the bearing of C from A is $110^\circ$. 8. Bearing of B from A. $\vec{AB}=B-A\approx(-0.060307,-0.342020)$. Compute $\arctan\left(\left|\dfrac{\Delta x}{\Delta y}\right|\right)=\arctan(0.176327)\approx10^\circ$. Since $\Delta y<0$ we add $180^\circ$ to get the bearing $180^\circ+10^\circ=190^\circ$. Thus the bearing of B from A is $190^\circ$. Final answers. (a) Bearing of C from A: $110^\circ$. (b) Bearing of B from A: $190^\circ$.