Numerical Problem: In $\triangle ABC$, side $AB = 7$ cm, side $AC = 9$ cm and $\angle BAC = 60^{\circ}$.\ a) Find the length of side $BC$ using the Cosine Rule.\ b) Calculate the area of the triangle using the formula $\text{Area}=\frac{1}{2}ab\sin C$. Show all steps clearly.

Q: Numerical Problem: In $\triangle ABC$, side $AB = 7$ cm, side $AC = 9$ cm and $\angle BAC = 60^{\circ}$.\ a) Find the length of side $BC$ using the Cosine Rule.\ b) Calculate the area of the triangle using the formula $\text{Area}=\frac{1}{2}ab\sin C$. Show all steps clearly.

a) Finding $BC$ using the Cosine Rule Identify the known values: $a = BC$ (the side we need to find) $b = AC = 9$ cm $c = AB = 7$ cm Included angle $A = \angle BAC = 60^{\circ}$. Write the Cosine Rule for side $a$: \a^{2}=b^{2}+c^{2}-2bc\cos A\ Substitute the known numbers: \a^{2}=9^{2}+7^{2}-2\times9\times7\cos 60^{\circ}\ \a^{2}=81+49-126\times\frac{1}{2}\ \a^{2}=130-63\ \a^{2}=67\ Take the square root to obtain $a$: \a=\sqrt{67}\approx 8.19\text{ cm}\ Hence, $BC \approx 8.19$ cm. b) Area of $\triangle ABC$ using $\frac{1}{2}ab\sin C$ Choose two sides and the included angle. We can use sides $AB$ and $AC$ with the included angle $A = 60^{\circ}$. Apply the area formula: \\text{Area}=\frac{1}{2}\times AB \times AC \times \sin 60^{\circ}\ \\text{Area}=\frac{1}{2}\times 7 \times 9 \times \frac{\sqrt{3}}{2}\ Calculate step‑by‑step: $\frac{1}{2}\times 7 \times 9 = \frac{63}{2}=31.5$ $\sin 60^{\circ}=\frac{\sqrt{3}}{2}\approx 0.866$ Multiply: $31.5 \times 0.866 \approx 27.27$ cm$^2$. Result: The area of $\triangle ABC$ is approximately $27.3$ square centimetres. Final Answers: $BC \approx 8.19$ cm Area $\approx 27.3$ cm$^2$

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Accepted Answer

a) Finding $BC$ using the Cosine Rule Identify the known values: $a = BC$ (the side we need to find) $b = AC = 9$ cm $c = AB = 7$ cm Included angle $A = \angle BAC = 60^{\circ}$. Write the Cosine Rule for side $a$: \a^{2}=b^{2}+c^{2}-2bc\cos A\ Substitute the known numbers: \a^{2}=9^{2}+7^{2}-2\times9\times7\cos 60^{\circ}\ \a^{2}=81+49-126\times\frac{1}{2}\ \a^{2}=130-63\ \a^{2}=67\ Take the square root to obtain $a$: \a=\sqrt{67}\approx 8.19\text{ cm}\ Hence, $BC \approx 8.19$ cm. b) Area of $\triangle ABC$ using $\frac{1}{2}ab\sin C$ Choose two sides and the included angle. We can use sides $AB$ and $AC$ with the included angle $A = 60^{\circ}$. Apply the area formula: \text{Area}=\frac{1}{2}\times AB \times AC \times \sin 60^{\circ}\ \text{Area}=\frac{1}{2}\times 7 \times 9 \times \frac{\sqrt{3}}{2}\ Calculate step‑by‑step: $\frac{1}{2}\times 7 \times 9 = \frac{63}{2}=31.5$ $\sin 60^{\circ}=\frac{\sqrt{3}}{2}\approx 0.866$ Multiply: $31.5 \times 0.866 \approx 27.27$ cm$^2$. Result: The area of $\triangle ABC$ is approximately $27.3$ square centimetres. Final Answers: $BC \approx 8.19$ cm Area $\approx 27.3$ cm$^2$

a) Finding (BC) using the Cosine Rule

b) Area of (\triangle ABC) using (\frac{1}{2}ab\sin C)