1. **Problem statement:** We are given bearings and travel times between three cities A, B, and C. The bearing from A to B is S 65° E, and from B to C is N 50° E. A car travels from A to B at 60 mph for 1.3 hours, and from B to C for 2.7 hours at the same speed. We need to find the distance between city A and city C. 2. **Calculate distances AB and BC:** - Distance = speed × time - $AB = 60 \times 1.3 = 78$ miles - $BC = 60 \times 2.7 = 162$ miles 3. **Determine the angle between paths AB and BC:** - Bearing from A to B is S 65° E, which means 65° east of south. - Bearing from B to C is N 50° E, which means 50° east of north. - The angle at B between AB and BC is $\theta = 65^\circ + 50^\circ = 115^\circ$ because the two bearings are on opposite sides of the north-south line. 4. **Use the Law of Cosines to find distance AC:** - The triangle formed is ABC with sides AB, BC, and AC opposite angle $\theta$. - Law of Cosines formula: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\theta)$$ - Substitute values: $$AC^2 = 78^2 + 162^2 - 2 \times 78 \times 162 \times \cos(115^\circ)$$ 5. **Calculate:** - $78^2 = 6084$ - $162^2 = 26244$ - $\cos(115^\circ) = \cos(180^\circ - 65^\circ) = -\cos(65^\circ) \approx -0.4226$ - So, $$AC^2 = 6084 + 26244 - 2 \times 78 \times 162 \times (-0.4226)$$ - Calculate the product: $$2 \times 78 \times 162 = 25272$$ - Then, $$AC^2 = 6084 + 26244 + 25272 \times 0.4226 = 32328 + 10675.5 = 43003.5$$ 6. **Find AC:** $$AC = \sqrt{43003.5} \approx 207.4$$ miles **Final answer:** The distance between city A and city C is approximately **207.4 miles**.