1. **Problem Statement:** We have a triangle with vertices T, B, and F. Given: - Distance TB = 130 m - Distance TF = 185 m - Angle $\angle BTF = 38^\circ$ - Angle $\angle TFB = 12^\circ$ We need to find: (i) Distance BF using the Cosine Rule. (ii) Angle $\angle TFB$ using the Sine Rule. (iii) Area of sector GFE of a circle with diameter 28 m and central angle $\angle TFB$. 2. **(i) Find |BF| using the Cosine Rule:** The Cosine Rule states: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ where $c$ is the side opposite angle $C$. Here, $a = TB = 130$, $b = TF = 185$, and $C = \angle BTF = 38^\circ$. Calculate: $$BF^2 = 130^2 + 185^2 - 2 \times 130 \times 185 \times \cos(38^\circ)$$ $$= 16900 + 34225 - 48100 \times \cos(38^\circ)$$ Calculate $\cos(38^\circ) \approx 0.7880$: $$BF^2 = 16900 + 34225 - 48100 \times 0.7880 = 51125 - 37902.8 = 13222.2$$ So, $$BF = \sqrt{13222.2} \approx 115\text{ m (nearest metre)}$$ 3. **(ii) Find $|\angle TFB|$ using the Sine Rule:** The Sine Rule states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ We know: - Side BF $= 115$ m (from part i) - Side TB $= 130$ m - Angle $\angle BTF = 38^\circ$ - Angle $\angle TFB = 12^\circ$ (given) We want to find $\angle TFB$ which is already given as 12°, so we find $\angle TFB$ using the Sine Rule to confirm or find $\angle TFB$ if needed. Using Sine Rule between sides TB and BF: $$\frac{BF}{\sin(\angle BTF)} = \frac{TB}{\sin(\angle TFB)}$$ Rearranged: $$\sin(\angle TFB) = \frac{TB \times \sin(\angle BTF)}{BF} = \frac{130 \times \sin(38^\circ)}{115}$$ Calculate $\sin(38^\circ) \approx 0.6157$: $$\sin(\angle TFB) = \frac{130 \times 0.6157}{115} = \frac{80.04}{115} \approx 0.696$$ So, $$\angle TFB = \arcsin(0.696) \approx 44^\circ$$ 4. **(iii) Find the area of sector GFE:** The flag F is the center of a circle with diameter 28 m, so radius $r = \frac{28}{2} = 14$ m. The sector angle is $\angle TFB = 44^\circ$ (from part ii). Area of sector formula: $$\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$$ Calculate: $$= \frac{44}{360} \times \pi \times 14^2 = \frac{44}{360} \times \pi \times 196$$ $$= 0.1222 \times 3.1416 \times 196 = 75.3\text{ m}^2$$ 5. **(b)(i) Maximum and minimum lengths of Rory's shot:** Rory's ball lands 10 m from F, so the shot length is the distance from T to P where P lies on a circle of radius 10 m centered at F. Minimum length = $TF - FP = 185 - 10 = 175$ m Maximum length = $TF + FP = 185 + 10 = 195$ m 6. **(b)(ii) Find $|\angle FTP|$ when Rory's shot length is 187 m:** Triangle TFP with sides: - TF = 185 m - FP = 10 m - TP = 187 m Use Cosine Rule to find $\angle FTP$: $$TP^2 = TF^2 + FP^2 - 2 \times TF \times FP \times \cos(\angle FTP)$$ Rearranged: $$\cos(\angle FTP) = \frac{TF^2 + FP^2 - TP^2}{2 \times TF \times FP} = \frac{185^2 + 10^2 - 187^2}{2 \times 185 \times 10}$$ Calculate: $$= \frac{34225 + 100 - 34969}{3700} = \frac{34225 + 100 - 34969}{3700} = \frac{-644}{3700} \approx -0.1741$$ So, $$\angle FTP = \arccos(-0.1741) \approx 100^\circ$$ **Final answers:** (i) $|BF| = 115$ m (ii) $|\angle TFB| = 44^\circ$ (iii) Area of sector GFE = 75.3 m² (b)(i) Minimum length = 175 m, Maximum length = 195 m (b)(ii) $|\angle FTP| = 100^\circ$