1. **Problem statement:** We have triangle ABC on horizontal ground with points A, B, and C. - Bearing of C from A is 145°. - Angle ACB = 38°. - AC = 65 m, BC = 95 m. We need to find: (a) Bearing of B from C. (b) Length AB. (c) Bearing of B from A, given angle BAC is obtuse. 2. **Key formulas and rules:** - Bearing is measured clockwise from North. - Use the Law of Cosines to find side AB: $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(\angle ACB)$$ - Use the Law of Sines or angle sum properties to find unknown angles. 3. **Step (a): Find bearing of B from C.** - Bearing of C from A is 145°, so line AC makes 145° clockwise from North at A. - Angle ACB = 38° is the angle at C between points A and B. - To find bearing of B from C, we find the direction of line CB relative to North. Since angle ACB = 38°, and AC is at bearing 145° from A, the angle at C between A and B is 38°, so the bearing of B from C is: $$\text{Bearing of B from C} = 145^\circ - 180^\circ + 38^\circ = 3^\circ$$ Explanation: - The line AC from A to C is 145°. - At C, angle between A and B is 38°, so B lies 38° clockwise from line CA (which is 325° from North at C because bearing from A to C is 145°, so from C to A is 145° - 180° = -35° or 325°). - Therefore, bearing of B from C is $325^\circ + 38^\circ = 363^\circ$, which is equivalent to $3^\circ$. 4. **Step (b): Find length AB using Law of Cosines.** $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(38^\circ)$$ $$= 65^2 + 95^2 - 2 \times 65 \times 95 \times \cos(38^\circ)$$ Calculate: $$65^2 = 4225$$ $$95^2 = 9025$$ $$2 \times 65 \times 95 = 12350$$ $$\cos(38^\circ) \approx 0.7880$$ So: $$AB^2 = 4225 + 9025 - 12350 \times 0.7880 = 13250 - 9731.8 = 3518.2$$ $$AB = \sqrt{3518.2} \approx 59.3$$ 5. **Step (c): Find bearing of B from A given angle BAC is obtuse.** - Angle BAC is obtuse, so angle at A is > 90°. - Use Law of Sines to find angle BAC: $$\frac{\sin(38^\circ)}{AB} = \frac{\sin(\angle BAC)}{BC}$$ $$\sin(\angle BAC) = \frac{BC \times \sin(38^\circ)}{AB} = \frac{95 \times 0.6157}{59.3} = 0.986$$ $$\angle BAC = \sin^{-1}(0.986) \approx 80.1^\circ$$ Since angle BAC is obtuse, actual angle is: $$180^\circ - 80.1^\circ = 99.9^\circ$$ - Bearing of C from A is 145°. - Angle BAC = 99.9°, so bearing of B from A is: $$145^\circ - 99.9^\circ = 45.1^\circ$$ **Final answers:** (a) Bearing of B from C is $3^\circ$. (b) Length AB is $59.3$ m. (c) Bearing of B from A is $45.1^\circ$.