1. **State the problem:** We need to find the distance between yachts B and C (denoted as $a = BC$) using the given triangle with sides $b = 500$ m, $c = 700$ m, and angle $A = 15^\circ$ between them. 2. **Formula used:** The Law of Cosines states: $$a^2 = b^2 + c^2 - 2bc \cos A$$ This formula relates the lengths of sides of a triangle to the cosine of one of its angles. 3. **Calculate $a = BC$:** $$a^2 = 500^2 + 700^2 - 2 \times 500 \times 700 \times \cos 15^\circ$$ Calculate each term: $$500^2 = 250000$$ $$700^2 = 490000$$ $$2 \times 500 \times 700 = 700000$$ $$\cos 15^\circ \approx 0.9659$$ So, $$a^2 = 250000 + 490000 - 700000 \times 0.9659$$ $$a^2 = 740000 - 676130 = 63870$$ 4. **Find $a$ by taking the square root:** $$a = \sqrt{63870} \approx 252.7$$ 5. **Answer for (a):** The distance between yachts B and C is approximately **253 metres** (to 3 significant figures). --- 6. **Calculate the bearing $\theta$ of yacht C from yacht B:** We use the Law of Sines: $$\frac{\sin A}{a} = \frac{\sin \theta}{c}$$ Rearranged: $$\sin \theta = \frac{c \sin A}{a} = \frac{700 \times \sin 15^\circ}{252.7}$$ Calculate $\sin 15^\circ \approx 0.2588$: $$\sin \theta = \frac{700 \times 0.2588}{252.7} = \frac{181.16}{252.7} \approx 0.7169$$ 7. **Find $\theta$:** $$\theta = \arcsin(0.7169) \approx 45.8^\circ$$ 8. **Answer for (b):** The bearing of yacht C from yacht B is approximately **45.8 degrees**. **Final answers:** - (a) Distance $BC \approx 253$ m - (b) Bearing $\theta \approx 45.8^\circ$