1. **State the problem:** We have triangle $\triangle ABC$ with sides $a = 15$ cm, $b = 18$ cm, and angle $\angle C = 48^\circ$. We want to find side $AB$ and angle $A$ using both the cosine law and sine law. 2. **Cosine Law for side $AB$ (side $c$):** The cosine law states: $$c^2 = a^2 + b^2 - 2ab \cos C$$ where $c = AB$, $a = 15$, $b = 18$, and $C = 48^\circ$. 3. **Calculate $c$ using cosine law:** $$c^2 = 15^2 + 18^2 - 2 \times 15 \times 18 \times \cos 48^\circ$$ $$c^2 = 225 + 324 - 540 \times \cos 48^\circ$$ Calculate $\cos 48^\circ \approx 0.6691$: $$c^2 = 549 - 540 \times 0.6691 = 549 - 361.314 = 187.686$$ $$c = \sqrt{187.686} \approx 13.7 \text{ cm}$$ 4. **Cosine Law for angle $A$:** Rearranged cosine law to find angle $A$: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ Substitute $a=15$, $b=18$, $c=13.7$: $$\cos A = \frac{18^2 + 13.7^2 - 15^2}{2 \times 18 \times 13.7} = \frac{324 + 187.69 - 225}{493.2} = \frac{286.69}{493.2} \approx 0.5815$$ $$A = \cos^{-1}(0.5815) \approx 54.5^\circ$$ 5. **Sine Law for side $AB$ (side $c$):** Sine law states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ We know $a=15$, $b=18$, $C=48^\circ$, and want $c$. $$c = \frac{\sin C}{\sin A} \times a$$ Using $A \approx 54.5^\circ$ and $C=48^\circ$: $$c = \frac{\sin 48^\circ}{\sin 54.5^\circ} \times 15$$ Calculate $\sin 48^\circ \approx 0.7431$, $\sin 54.5^\circ \approx 0.8130$: $$c = \frac{0.7431}{0.8130} \times 15 = 0.914 \times 15 = 13.7 \text{ cm}$$ 6. **Sine Law for angle $A$:** Rearranged sine law: $$\sin A = \frac{a}{c} \times \sin C$$ Substitute $a=15$, $c=13.7$, $C=48^\circ$: $$\sin A = \frac{15}{13.7} \times 0.7431 = 1.0949 \times 0.7431 = 0.813$$ $$A = \sin^{-1}(0.813) \approx 54.5^\circ$$ **Final answers:** - Side $AB \approx 13.7$ cm - Angle $A \approx 54.5^\circ$ These results are consistent using both cosine and sine laws.