1. **Problem Statement:** (a) Find the size of angle $\angle ACB$ in triangle $ABC$ where $BC=AC=7$ cm. (b) Find the size of angle $\angle CAD$ in triangle $ABD$ with $AB=13$ cm, $AD=6.5$ cm, and $C$ on $BD$. 2. **Given:** - $AB=13$ cm, $AD=6.5$ cm - $BC=AC=7$ cm (isosceles triangle $ABC$) - $C$ lies on $BD$ 3. **Step (a): Find $\angle ACB$** - Since $BC=AC=7$ cm, triangle $ABC$ is isosceles with $AB$ as the base. - The angles opposite equal sides are equal, so $\angle ABC = \angle BAC$. - Use the Law of Cosines on triangle $ABC$ to find $\angle ACB$: $$AB^2 = AC^2 + BC^2 - 2 \times AC \times BC \times \cos(\angle ACB)$$ Substitute values: $$13^2 = 7^2 + 7^2 - 2 \times 7 \times 7 \times \cos(\angle ACB)$$ $$169 = 49 + 49 - 98 \cos(\angle ACB)$$ $$169 = 98 - 98 \cos(\angle ACB)$$ Rearranged: $$98 \cos(\angle ACB) = 98 - 169 = -71$$ $$\cos(\angle ACB) = \frac{-71}{98} \approx -0.7245$$ - Find $\angle ACB$: $$\angle ACB = \cos^{-1}(-0.7245) \approx 136.5^\circ$$ 4. **Step (b): Find $\angle CAD$** - Triangle $ABD$ with $AB=13$ cm, $AD=6.5$ cm. - Point $C$ lies on $BD$ such that $BC=AC=7$ cm. - Use the Law of Cosines in triangle $ABC$ to find $\angle BAC$ (which is $\angle CAD$ since $C$ lies on $BD$): $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle BAC)$$ Substitute values: $$7^2 = 13^2 + 7^2 - 2 \times 13 \times 7 \times \cos(\angle BAC)$$ $$49 = 169 + 49 - 182 \cos(\angle BAC)$$ $$49 = 218 - 182 \cos(\angle BAC)$$ Rearranged: $$182 \cos(\angle BAC) = 218 - 49 = 169$$ $$\cos(\angle BAC) = \frac{169}{182} \approx 0.9297$$ - Find $\angle BAC$: $$\angle BAC = \cos^{-1}(0.9297) \approx 21.5^\circ$$ 5. **Problem 2:** Triangle $PQR$ with $PQ=7$ cm, $PR=10$ cm, and $\angle PQR=75^\circ$. 6. **Step (a): Find $\angle QPR$** - Use Law of Cosines in triangle $PQR$ to find $\angle QPR$ opposite side $QR$. - First find length $QR$ using Law of Cosines: $$QR^2 = PQ^2 + PR^2 - 2 \times PQ \times PR \times \cos(\angle PQR)$$ $$QR^2 = 7^2 + 10^2 - 2 \times 7 \times 10 \times \cos(75^\circ)$$ $$QR^2 = 49 + 100 - 140 \times 0.2588 = 149 - 36.23 = 112.77$$ $$QR = \sqrt{112.77} \approx 10.62 \text{ cm}$$ - Now find $\angle QPR$ opposite side $QR$: $$\cos(\angle QPR) = \frac{PQ^2 + PR^2 - QR^2}{2 \times PQ \times PR} = \frac{49 + 100 - 112.77}{2 \times 7 \times 10} = \frac{36.23}{140} = 0.2588$$ $$\angle QPR = \cos^{-1}(0.2588) = 75^\circ$$ 7. **Step (b): Find area of triangle $PQR$** - Use formula for area with two sides and included angle: $$\text{Area} = \frac{1}{2} \times PQ \times PR \times \sin(\angle PQR)$$ $$= \frac{1}{2} \times 7 \times 10 \times \sin(75^\circ)$$ $$= 35 \times 0.9659 = 33.81 \text{ cm}^2$$ **Final answers:** - (a) $\angle ACB \approx 136.5^\circ$ - (b) $\angle CAD \approx 21.5^\circ$ - (5a) $\angle QPR = 75^\circ$ - (5b) Area $\approx 33.81$ cm$^2$