1. **Problem statement:** We have four towns A, B, C, and D with given positions relative to A: - B is 25 km due East of A. - C is 25 km due North of A. - D is 45 km due South of A. We need to find: (a) The bearing of B from C. (b) The bearing of D from B. 2. **Set up coordinates:** Let A be at the origin $(0,0)$. - B is 25 km East: coordinates $(25,0)$. - C is 25 km North: coordinates $(0,25)$. - D is 45 km South: coordinates $(0,-45)$. 3. **(a) Bearing of B from C:** - Vector from C to B is $\overrightarrow{CB} = (25 - 0, 0 - 25) = (25, -25)$. - The angle $\theta$ from North (positive y-axis) clockwise to vector $\overrightarrow{CB}$ is given by: $$\theta = \arctan\left(\frac{|x|}{|y|}\right) = \arctan\left(\frac{25}{25}\right) = \arctan(1) = 45^\circ$$ - Since $x$ is positive and $y$ is negative, the vector points Southeast from C. - Bearing is measured clockwise from North, so bearing of B from C is: $$\boxed{135^\circ}$$ 4. **(b) Bearing of D from B:** - Vector from B to D is $\overrightarrow{BD} = (0 - 25, -45 - 0) = (-25, -45)$. - Calculate the angle $\phi$ between vector $\overrightarrow{BD}$ and North (positive y-axis): - First find the angle $\alpha$ the vector makes with the positive x-axis: $$\alpha = \arctan\left(\frac{|-45|}{|-25|}\right) = \arctan\left(\frac{45}{25}\right) = \arctan(1.8) \approx 61.93^\circ$$ - Since $x$ is negative and $y$ is negative, the vector lies in the third quadrant. - The angle from positive x-axis is $180^\circ + 61.93^\circ = 241.93^\circ$. - Bearing is measured clockwise from North (positive y-axis). North corresponds to $90^\circ$ from positive x-axis. - Bearing $= \alpha_{bearing} = 241.93^\circ - 90^\circ = 151.93^\circ$. - Rounded to nearest degree: $$\boxed{152^\circ}$$ **Final answers:** (a) $135^\circ$ (b) $152^\circ$