1. **Problem statement:** Calculate angles ACB and ACD, and the area of quadrilateral ABCD given sides AB=11 cm, BC=4 cm, AC=13 cm, DC=8 cm, and angle at D is 80°. 2. **Calculate angle ACB:** Use the Law of Cosines in triangle ABC: $$\cos(\angle ACB) = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC}$$ Substitute values: $$\cos(\angle ACB) = \frac{11^2 + 4^2 - 13^2}{2 \times 11 \times 4} = \frac{121 + 16 - 169}{88} = \frac{-32}{88} = -\frac{8}{22} = -0.3636$$ Calculate angle: $$\angle ACB = \cos^{-1}(-0.3636) \approx 111.3^\circ$$ 3. **Calculate angle ACD:** Triangle ACD has sides AC=13 cm, DC=8 cm, and angle at D is 80°. Use Law of Cosines to find side AD: $$AD^2 = AC^2 + DC^2 - 2 \times AC \times DC \times \cos(80^\circ)$$ Calculate: $$AD^2 = 13^2 + 8^2 - 2 \times 13 \times 8 \times \cos(80^\circ) = 169 + 64 - 208 \times 0.1736 = 233 - 36.13 = 196.87$$ $$AD = \sqrt{196.87} \approx 14.03 \text{ cm}$$ Now use Law of Cosines in triangle ACD to find angle ACD opposite side DC: $$\cos(\angle ACD) = \frac{AD^2 + AC^2 - DC^2}{2 \times AD \times AC} = \frac{14.03^2 + 13^2 - 8^2}{2 \times 14.03 \times 13} = \frac{196.87 + 169 - 64}{364.78} = \frac{301.87}{364.78} = 0.8277$$ Calculate angle: $$\angle ACD = \cos^{-1}(0.8277) \approx 34.1^\circ$$ 4. **Calculate area of quadrilateral ABCD:** Split quadrilateral into triangles ABC and ACD. Area of triangle ABC using Heron's formula: $$s = \frac{11 + 4 + 13}{2} = 14$$ $$\text{Area}_{ABC} = \sqrt{s(s-11)(s-4)(s-13)} = \sqrt{14 \times 3 \times 10 \times 1} = \sqrt{420} \approx 20.49 \text{ cm}^2$$ Area of triangle ACD using formula with included angle 80°: $$\text{Area}_{ACD} = \frac{1}{2} \times AC \times DC \times \sin(80^\circ) = 0.5 \times 13 \times 8 \times 0.9848 = 51.1 \text{ cm}^2$$ Total area: $$\text{Area}_{ABCD} = 20.49 + 51.1 = 71.59 \text{ cm}^2$$ --- 5. **Sequences table completion:** - Sequence A: 2, 5, 8, 11, ... (arithmetic, difference 3) - Next term: $11 + 3 = 14$ - $n$th term: $a_n = 2 + (n-1) \times 3 = 3n - 1$ - Sequence B: 20, 14, 8, 2, ... (arithmetic, difference -6) - Next term: $2 - 6 = -4$ - $n$th term: $a_n = 20 + (n-1)(-6) = 26 - 6n$ - Sequence C: 1, 4, 9, 16, ... (perfect squares) - Next term: $25$ - $n$th term: $a_n = n^2$ - Sequence D: 0, 2, 6, 12, ... (pattern of $n(n-1)$) - Next term: $20$ (since $5 \times 4 = 20$) - $n$th term: $a_n = n(n-1)$ 6. **Sum of first n terms:** Given sum: $$S_n = \frac{n(3n + 1)}{2}$$ When $S_n = 155$: $$\frac{n(3n + 1)}{2} = 155 \Rightarrow n(3n + 1) = 310 \Rightarrow 3n^2 + n - 310 = 0$$ This confirms the quadratic equation.