1. **Stating the problem:** We have three power stations K, L, and M with the following information: - K is 60 km due north of L. - M is on a bearing of 130° from K. - M is on a bearing of 070° from L. We need to: (a) Construct a scale diagram with 1 cm representing 10 km. (b) Measure the actual distance between M and K. (c) Find the bearing of L from M. --- 2. **Constructing the scale diagram:** - Since 1 cm represents 10 km, 60 km north from L to K corresponds to 6 cm straight up on the diagram. - Place point L at the given position. - From L, draw a vertical line upwards 6 cm to mark K. 3. **Locating point M using bearings:** - From K, draw a line at a 130° bearing (measured clockwise from north). - From L, draw a line at a 070° bearing. - The intersection of these two lines locates M. 4. **Calculate actual distance MK (Answer b):** Bearings give directions: - K is at (0,60) relative to L at origin (0,0) in km. From K (0,60), line at 130° means direction vector: $$ (\sin 130^\circ, \cos 130^\circ) = (\sin 130^\circ, \cos 130^\circ) $$ From L at (0,0), line at 70° means direction vector: $$ (\sin 70^\circ, \cos 70^\circ) $$ Let the point M be $M = K + t(\sin 130^\circ, \cos 130^\circ)$ and also $M = L + s(\sin 70^\circ, \cos 70^\circ)$ for some scalars $t, s > 0$. Therefore: $$ (0,60) + t(\sin 130^\circ, \cos 130^\circ) = s(\sin 70^\circ, \cos 70^\circ) $$ which gives system: $$ t \sin 130^\circ = s \sin 70^\circ $$ $$ 60 + t \cos 130^\circ = s \cos 70^\circ $$ Calculate approximate sine and cosine values: $$ \sin 130^\circ \approx 0.7660 $$ $$ \cos 130^\circ \approx -0.6428 $$ $$ \sin 70^\circ \approx 0.9397 $$ $$ \cos 70^\circ \approx 0.3420 $$ From first equation: $$ t \times 0.7660 = s \times 0.9397 \Rightarrow s = \frac{0.7660}{0.9397} t \approx 0.815 t $$ Substitute in second equation: $$ 60 - 0.6428 t = 0.3420 s = 0.3420 \times 0.815 t = 0.279 t $$ So: $$ 60 = 0.279 t + 0.6428 t = 0.9218 t \Rightarrow t = \frac{60}{0.9218} \approx 65.08 $$ Distance $MK = t \times$ magnitude of unit vector in that direction = $t \times 1 = 65.08$ km. **Answer (b):** The actual distance from M to K is approximately 65.1 km. 5. **Find bearing of L from M (Answer c):** Coordinates of M: $$ M = K + t(\sin 130^\circ, \cos 130^\circ) = (0,60) + 65.08 \times (0.7660, -0.6428) = (49.8, 18.2) $$ Vector from M to L: $$ \overrightarrow{ML} = L - M = (0 - 49.8, 0 - 18.2) = (-49.8, -18.2) $$ Calculate angle $\theta$ from north to $\overrightarrow{ML}$: - Angle with positive y-axis (north) is: $$ \theta = \arctan \left( \frac{|x|}{|y|} \right) = \arctan \left( \frac{49.8}{18.2} \right) \approx 69.6^\circ $$ Since vector points southwest (x negative, y negative), bearing from M to L is: $$ 180^\circ + 69.6^\circ = 249.6^\circ $$ **Answer (c):** Bearing of L from M is approximately $250^\circ$.