1. **Problem 24:** Find the longest side in triangle XYZ with angles 40° at X, 60° at Z, and side XZ = 12.0 cm. 2. **Recall:** The longest side is opposite the largest angle. 3. Angles: $\angle X = 40^\circ$, $\angle Z = 60^\circ$, so $\angle Y = 180^\circ - 40^\circ - 60^\circ = 80^\circ$. 4. Largest angle is $80^\circ$ at Y, so side opposite Y is XZ = 12.0 cm. 5. Sides XY and YZ are marked as $x$; use Law of Sines: $$\frac{XY}{\sin 60^\circ} = \frac{YZ}{\sin 40^\circ} = \frac{XZ}{\sin 80^\circ}$$ 6. Calculate $x = XY = YZ$: $$x = \frac{XZ \times \sin 60^\circ}{\sin 80^\circ} = \frac{12.0 \times \sin 60^\circ}{\sin 80^\circ}$$ 7. Substitute values: $$\sin 60^\circ \approx 0.866, \quad \sin 80^\circ \approx 0.9848$$ 8. Compute: $$x = \frac{12.0 \times 0.866}{0.9848} \approx \frac{10.392}{0.9848} \approx 10.55 \text{ cm}$$ 9. Compare sides: - $XY = YZ \approx 10.55$ cm - $XZ = 12.0$ cm 10. Longest side is $XZ$. --- 11. **Problem 25:** The Pythagorean Theorem, Sine Law, Cosine Law, Tangent Law, and Primary Trigonometric Ratios are all topics studied in the first five lessons. 12. Answer: a) True. --- 13. **Problem 26a:** Given $\triangle ABC \cong \triangle ZYX$, find side $a$ (corresponding to side AC = 1.96 cm). 14. Since congruent triangles have equal corresponding sides, $a = 1.96$ cm. 15. **Problem 26b:** Given $\triangle ABC \sim \triangle WVU$, find side $u$. 16. Similar triangles have proportional sides: $$\frac{AB}{WV} = \frac{AC}{WU} = \frac{BC}{VU}$$ 17. Given $AB=2.35$ cm, $WV=2.94$ cm, $AC=1.96$ cm, $WU=37^\circ$ (angle, so use side opposite angle), find $u$ corresponding to side $AC$ scaled. 18. Use ratio: $$\frac{u}{1.96} = \frac{2.94}{2.35} \Rightarrow u = 1.96 \times \frac{2.94}{2.35}$$ 19. Calculate: $$u \approx 1.96 \times 1.251 = 2.45 \text{ cm}$$ --- 20. **Problem 27a:** Find height $h$ of flagpole using similar triangles. 21. Ratios: $$\frac{h}{14} = \frac{1}{2}$$ 22. Solve: $$h = \frac{14}{2} = 7$$ 23. Height of flagpole is 7 m. 24. **Problem 27b:** Find hypotenuse $c$ of triangle with height $h=7$ m and base 14 m. 25. Use Pythagorean theorem: $$c = \sqrt{h^2 + 14^2} = \sqrt{7^2 + 14^2} = \sqrt{49 + 196} = \sqrt{245}$$ 26. Calculate: $$c \approx 15.65 \text{ m}$$ --- 27. **Problem 28:** Solve for $\angle C$ in $\triangle ABC$ with $AB=10$, $BC=12$, $\angle A=75^\circ$. 28. Use Law of Cosines or Law of Sines. Use Law of Sines: $$\frac{\sin C}{AB} = \frac{\sin A}{BC} \Rightarrow \sin C = \frac{AB}{BC} \sin A = \frac{10}{12} \times \sin 75^\circ$$ 29. Calculate: $$\sin 75^\circ \approx 0.9659$$ 30. So: $$\sin C = \frac{10}{12} \times 0.9659 = 0.805$$ 31. Find $C$: $$C = \sin^{-1}(0.805) \approx 53.5^\circ$$ --- 32. **Problem 29a:** Calculate height $h$ of clouds with angle 30° and distance $d=300$ m. 33. Use: $$\tan 30^\circ = \frac{h}{300} \Rightarrow h = 300 \times \tan 30^\circ$$ 34. Calculate: $$\tan 30^\circ \approx 0.5774$$ 35. So: $$h = 300 \times 0.5774 = 173.2 \text{ m}$$ --- 36. **Problem 29b:** Given $h=500$ m and $d=300$ m, find angle $\theta$. 37. Use: $$\tan \theta = \frac{h}{d} = \frac{500}{300} = 1.6667$$ 38. Find $\theta$: $$\theta = \tan^{-1}(1.6667) \approx 59.0^\circ$$ --- 39. **Problem 30:** Solve for $x$ in triangle with sides 8 cm, 11 cm, and angle 75° opposite $x$. 40. Use Law of Cosines: $$x^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos 75^\circ$$ 41. Calculate: $$\cos 75^\circ \approx 0.2588$$ 42. So: $$x^2 = 64 + 121 - 176 \times 0.2588 = 185 - 45.53 = 139.47$$ 43. Find $x$: $$x = \sqrt{139.47} \approx 11.81 \text{ cm}$$ --- **Final answers:** - 24: Longest side is XZ (12.0 cm) - 25: a) True - 26a: $a = 1.96$ cm - 26b: $u \approx 2.45$ cm - 27a: Height $h = 7$ m - 27b: Hypotenuse $c \approx 15.65$ m - 28: $\angle C \approx 53.5^\circ$ - 29a: Height $h \approx 173.2$ m - 29b: Angle $\theta \approx 59.0^\circ$ - 30: $x \approx 11.81$ cm