1. **Problem statement:** We have a shape formed by joining triangle ABC and sector CBD of a circle with radius 7 cm and center C. Points A, C, and D lie on a straight line with AC = 9 cm and CD = 7 cm. Angle BAC = 0.5 radians. We need to find: a) The angle ACB in radians. b) The area of the shape ABDC. --- 2. **Given data:** - AC = 9 cm - CD = 7 cm (radius of the circle) - CB = 7 cm (radius of the circle) - Angle BAC = 0.5 radians --- 3. **Step a) Find angle ACB:** - Triangle ABC has points A, B, C. - We know AC = 9 cm. - CB = 7 cm. - Angle BAC = 0.5 radians. Use the Law of Cosines in triangle ABC to find side AB: $$AB^2 = AC^2 + CB^2 - 2 \times AC \times CB \times \cos(\angle BAC)$$ Calculate: $$AB^2 = 9^2 + 7^2 - 2 \times 9 \times 7 \times \cos(0.5)$$ $$= 81 + 49 - 126 \times \cos(0.5)$$ Calculate $\cos(0.5)$ approximately: $$\cos(0.5) \approx 0.87758$$ So: $$AB^2 = 130 - 126 \times 0.87758 = 130 - 110.56 = 19.44$$ Therefore: $$AB = \sqrt{19.44} \approx 4.41 \text{ cm}$$ --- Now, find angle ACB using the Law of Cosines again in triangle ABC: $$\cos(\angle ACB) = \frac{AC^2 + CB^2 - AB^2}{2 \times AC \times CB}$$ Substitute values: $$= \frac{9^2 + 7^2 - 4.41^2}{2 \times 9 \times 7} = \frac{81 + 49 - 19.44}{126} = \frac{110.56}{126} \approx 0.87758$$ So: $$\angle ACB = \cos^{-1}(0.87758) = 0.5 \text{ radians}$$ --- 4. **Step b) Find area of ABDC:** The shape ABDC consists of triangle ABC plus sector CBD. - Area of triangle ABC: Use formula: $$\text{Area} = \frac{1}{2} \times AC \times AB \times \sin(\angle BAC)$$ Calculate $\sin(0.5)$: $$\sin(0.5) \approx 0.47943$$ So: $$\text{Area}_{ABC} = \frac{1}{2} \times 9 \times 4.41 \times 0.47943 \approx 9.5 \text{ cm}^2$$ - Area of sector CBD: Radius $r = 7$ cm. Angle at center C is angle BCD. Since points C, D, A are collinear and CD = 7 cm, the sector angle is the same as angle ACB which is 0.5 radians. Area of sector: $$\text{Area}_{sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 7^2 \times 0.5 = \frac{1}{2} \times 49 \times 0.5 = 12.25 \text{ cm}^2$$ --- 5. **Total area ABDC:** $$\text{Area}_{ABDC} = \text{Area}_{ABC} + \text{Area}_{sector} = 9.5 + 12.25 = 21.75 \text{ cm}^2$$ --- **Final answers:** a) Angle ACB = $0.5$ radians b) Area ABDC = $21.75$ cm$^2$