1. **Problem statement:** A cyclist starts from point X, rides 3 km due west to point A, then rides 5 km northwest to point Z. We need to find: (i) The distance from the starting point X to point Z. (ii) The bearing of Z from X. 2. **Understanding directions:** - West means moving along the negative x-axis. - Northwest means moving at a 45° angle between north and west. 3. **Step 1: Represent the movements as vectors.** - From X to A: 3 km west means vector $ \vec{XA} = (-3, 0) $. - From A to Z: 5 km northwest means 5 km at 135° from the positive x-axis (since 0° is east, 90° is north, 135° is northwest). Calculate components of $ \vec{AZ} $: $$ \vec{AZ}_x = 5 \cos 135^\circ = 5 \times \left(-\frac{\sqrt{2}}{2}\right) = -\frac{5\sqrt{2}}{2} $$ $$ \vec{AZ}_y = 5 \sin 135^\circ = 5 \times \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2} $$ 4. **Step 2: Find the position vector of Z relative to X.** $$ \vec{XZ} = \vec{XA} + \vec{AZ} = (-3, 0) + \left(-\frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2}\right) = \left(-3 - \frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2}\right) $$ 5. **Step 3: Calculate the distance from X to Z.** $$ \text{Distance} = \sqrt{\left(-3 - \frac{5\sqrt{2}}{2}\right)^2 + \left(\frac{5\sqrt{2}}{2}\right)^2} $$ First, compute each term: $$ \left(-3 - \frac{5\sqrt{2}}{2}\right)^2 = \left(-3 - 3.5355\right)^2 = (-6.5355)^2 = 42.71 $$ $$ \left(\frac{5\sqrt{2}}{2}\right)^2 = (3.5355)^2 = 12.5 $$ Sum: $$ 42.71 + 12.5 = 55.21 $$ Distance: $$ \sqrt{55.21} \approx 7.43 \text{ km} $$ 6. **Step 4: Find the bearing of Z from X.** Bearing is measured clockwise from north. Calculate angle $ \theta $ from the positive x-axis: $$ \theta = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{\frac{5\sqrt{2}}{2}}{-3 - \frac{5\sqrt{2}}{2}} \right) = \tan^{-1} \left( \frac{3.5355}{-6.5355} \right) = \tan^{-1}(-0.54) = -28.3^\circ $$ Since x is negative and y is positive, point Z is in the second quadrant, so angle from positive x-axis is: $$ 180^\circ - 28.3^\circ = 151.7^\circ $$ Bearing from north (0°) clockwise is: $$ \text{Bearing} = 360^\circ - (151.7^\circ - 90^\circ) = 360^\circ - 61.7^\circ = 298.3^\circ $$ Alternatively, bearing from north is: $$ 90^\circ + 28.3^\circ = 118.3^\circ $$ But since bearing is clockwise from north, and the point is northwest, the bearing is: $$ 360^\circ - 118.3^\circ = 241.7^\circ $$ The correct bearing is approximately $ 241.7^\circ $ (west of north). **Final answers:** (i) Distance from X to Z is approximately **7.43 km**. (ii) Bearing of Z from X is approximately **242°** (rounded).