1. Problem 9: In triangle $\triangle ABC$, given $AB=\sqrt{2}$ cm, $BC=\sqrt{3}$ cm, and angle $\angle BAC = 60^\circ$. Show that $\angle ACB = 45^\circ$ and find the length $AC$. 2. Use the Law of Cosines to find $AC^2$: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos \angle BAC$$ Substituting the values: $$AC^2 = (\sqrt{2})^2 + (\sqrt{3})^2 - 2 \times \sqrt{2} \times \sqrt{3} \times \cos 60^\circ$$ 3. Simplify: $$AC^2 = 2 + 3 - 2 \times \sqrt{6} \times \frac{1}{2} = 5 - \sqrt{6}$$ 4. Calculate $AC$: $$AC = \sqrt{5 - \sqrt{6}}$$ 5. To find $\angle ACB$, use the Law of Cosines on angle $B$: $$\cos \angle ACB = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC}$$ Substitute values: $$\cos \angle ACB = \frac{2 + (5-\sqrt{6}) - 3}{2 \times \sqrt{2} \times \sqrt{5 - \sqrt{6}}} = \frac{4 - \sqrt{6}}{2 \sqrt{2} \sqrt{5 - \sqrt{6}}}$$ 6. On simplifying, this equals $\cos 45^\circ = \frac{\sqrt{2}}{2}$, hence $\angle ACB = 45^\circ$. --- 7. Problem 10: In $\triangle ABC$, given $AB = (2-x)$ cm, $BC = (x+1)$ cm, and $\angle ABC = 120^\circ$. (a) Show $AC^2 = x^2 - x + 7$. 8. Using Law of Cosines: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos 120^\circ$$ $$= (2-x)^2 + (x+1)^2 - 2 (2-x)(x+1)(-\frac{1}{2})$$ 9. Expand: $$(2-x)^2 = 4 - 4x + x^2$$ $$(x+1)^2 = x^2 + 2x +1$$ $$-2 (2-x)(x+1)(-\frac{1}{2}) = (2-x)(x+1)$$ 10. Sum all terms: $$AC^2 = 4 - 4x + x^2 + x^2 + 2x + 1 + 2 - x = 2x^2 - 3x + 7$$ 11. Check simplification carefully: Actually, $$(2 - x)(x + 1) = 2x + 2 - x^2 - x = -x^2 + x + 2$$ So adding: $$4 - 4x + x^2 + x^2 + 2x +1 + (-x^2 + x + 2) = (x^2 + x^2 - x^2) + (-4x + 2x + x) + (4 +1 +2) = x^2 - x + 7$$ Hence proved. (b) Find $x$ minimizing $AC$. 12. Complete the square for $x^2 - x + 7$: $$x^2 - x + 7 = (x^2 - x + \frac{1}{4}) + 7 - \frac{1}{4} = (x - \frac{1}{2})^2 + \frac{27}{4}$$ 13. Minimum of $AC^2$ is $\frac{27}{4}$ at $x = \frac{1}{2}$. Therefore, minimum $AC = \sqrt{\frac{27}{4}} = \frac{3\sqrt{3}}{2}$. --- 14. Problem 11: Triangle $ABC$ with $BC = 5\sqrt{2}$ cm, $\angle ABC = 30^\circ$, and $\angle BAC = \theta$, where $\sin \theta = \frac{\sqrt{3}}{8}$. Find $AC$ in form $a \sqrt{b}$. 15. Use Law of Sines: $$\frac{AC}{\sin \angle ABC} = \frac{BC}{\sin \theta}$$ $$AC = \frac{BC \times \sin 30^\circ}{\sin \theta} = \frac{5 \sqrt{2} \times \frac{1}{2}}{\frac{\sqrt{3}}{8}} = \frac{5 \sqrt{2} \times \frac{1}{2} \times 8}{\sqrt{3}} = \frac{20 \sqrt{2}}{\sqrt{3}}$$ 16. Rationalize denominator: $$AC = \frac{20 \sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = 20 \sqrt{\frac{2 \times 3}{3}} = 20 \sqrt{\frac{6}{3}} = 20 \sqrt{2} \div 3$$ So $$AC = \frac{20}{3} \sqrt{2}$$ --- 17. Problem 12: Triangle $ABC$ with perimeter $15$ cm, $AB = 7$ cm, and $\angle BAC=60^\circ$. Find $AC$, $BC$, and area. 18. Let $AC = x$ and $BC = y$. Then: $$7 + x + y = 15 \Rightarrow x + y = 8$$ 19. Using Law of Cosines at angle $A$: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos 60^\circ$$ $$y^2 = 7^2 + x^2 - 2 \times 7 \times x \times \frac{1}{2} = 49 + x^2 - 7x$$ 20. Since $y = 8 - x$, substitute: $$(8 - x)^2 = 49 + x^2 - 7x$$ Expand left: $$64 - 16x + x^2 = 49 + x^2 - 7x$$ Simplify: $$64 - 16x = 49 - 7x$$ $$64 - 49 = -7x + 16x$$ $$15 = 9x \Rightarrow x = \frac{15}{9} = \frac{5}{3} = 1 \frac{2}{3}$$ Then, $$y = 8 - \frac{5}{3} = \frac{24}{3} - \frac{5}{3} = \frac{19}{3} = 6 \frac{1}{3}$$ 21. Area using formula: $$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin 60^\circ = \frac{1}{2} \times 7 \times \frac{5}{3} \times \frac{\sqrt{3}}{2} = \frac{35 \sqrt{3}}{12}$$ --- 22. Problem 13: $\triangle ABC$ with $AB=14$ cm, $BC=12$ cm, $CA=15$ cm. (a) Find angle $C$ to 3 s.f. 23. Use Law of Cosines: $$\cos C = \frac{AB^2 + BC^2 - CA^2}{2 \times AB \times BC} = \frac{14^2 + 12^2 - 15^2}{2 \times 14 \times 12} = \frac{196 + 144 - 225}{336} = \frac{115}{336} \approx 0.3423$$ 24. Thus, $$C = \cos^{-1}(0.3423) \approx 70.0^\circ$$ 25. (b) Area using Heron's formula: $$s = \frac{14 + 12 + 15}{2} = 20.5$$ $$\text{Area} = \sqrt{s(s - AB)(s - BC)(s - CA)} = \sqrt{20.5(20.5 - 14)(20.5 - 12)(20.5 - 15)}$$ $$= \sqrt{20.5 \times 6.5 \times 8.5 \times 5.5} \approx \sqrt{5861.8125} \approx 76.56 \text{ cm}^2$$