1. **Problem (a):** Calculate angle CÂB in the triangle ABC where AB=65 m and CB=200 m. 2. The triangle has points A (top of tower), B (base of tower), and C (car position). AB is vertical, CB horizontal. 3. To find angle CÂB, we use the cosine rule in triangle ABC: $$\cos(CÂB)=\frac{AB^2 + CB^2 - AC^2}{2 \cdot AB \cdot CB}$$ 4. First, find length AC using Pythagoras theorem because angle at B is right angled (vertical tower and horizontal ground): $$AC = \sqrt{AB^2 + CB^2} = \sqrt{65^2 + 200^2} = \sqrt{4225 + 40000} = \sqrt{44225}$$ 5. Calculate AC: $$AC = 210.29 \text{ m (approx)}$$ 6. Now use cosine rule to calculate angle CÂB: $$\cos(CÂB) = \frac{65^2 + 200^2 - 210.29^2}{2 \times 65 \times 200} = \frac{4225 + 40000 - 44225}{26000} = \frac{0}{26000} = 0$$ 7. Therefore, $$CÂB = \cos^{-1}(0) = 90^\circ$$ --- 8. **Problem (b)(i):** Boat travels from R to S at speed 5 km/h, RP = 300 m, RS = 750 m, R is north of P, S east of P. Find time to reach S from R. 9. Find distance RS using Pythagoras theorem since R is north and S is east of P: $$RS = \sqrt{RP^2 + PS^2}$$ PS = distance from P to S = 750 m 10. Calculate RS: $$RS = \sqrt{300^2 + 750^2} = \sqrt{90000 + 562500} = \sqrt{652500} = 807.79 \text{ m}$$ 11. Convert distance RS to km: $$RS = 0.80779 \text{ km}$$ 12. Time to travel = distance / speed: $$t = \frac{0.80779}{5} = 0.16156 \text{ hours}$$ 13. Convert hours to minutes: $$0.16156 \times 60 = 9.69 \text{ minutes} \approx 9 \text{ minutes} 41 \text{ seconds}$$ 14. Departure time from R is 22:56, so arrival time at S is: $$22:56 + 0:09:41 = 23:05:41$$ --- 15. **Problem (b)(ii):** Calculate the bearing of S from R. 16. The bearing is the clockwise angle from north line at R to the line RS. 17. Using trigonometry, tan of angle $\theta$ at R (bearing) is opposite over adjacent: $$\tan(\theta) = \frac{PS}{RP} = \frac{750}{300} = 2.5$$ 18. Calculate $\theta$: $$\theta = \tan^{-1}(2.5) = 68.2^\circ$$ 19. Bearing from north clockwise to S from R: $$\boxed{068^\circ}$$ **Final answers:** - (a) $CÂB = 90^\circ$ - (b)(i) Arrival time at S is approximately 23:05:41 - (b)(ii) Bearing of S from R is $68^\circ$