In triangle \(ABC\), \(AB = 8\) cm, \(AC = 5\) cm and angle \(BAC = 60^\circ\). (a) Calculate \(BC\). [3] (b) Work out the exact area of triangle \(ABC\). [...

(a) Two sides and the angle between them are given, which is exactly the situation for the cosine rule: \(a^{2}=b^{2}+c^{2}-2bc\cos A\), where \(A\) is the included angle and \(a\) the side opposite it.

1. \(BC^{2}=8^{2}+5^{2}-2\times 8\times 5\times\cos 60^{\circ}\) [M1].
2. Using the exact value \(\cos 60^{\circ}=\dfrac{1}{2}\), the last term is \(80\times\dfrac{1}{2}=40\), so \(BC^{2}=64+25-40=89-40=49\) [M1].
3. \(BC=\sqrt{49}=7\) cm [A1].

(b) With the same two sides and included angle, the area is \(\dfrac{1}{2}ab\sin C\):

Area \(=\dfrac{1}{2}\times 8\times 5\times\sin 60^{\circ}\) [M1] \(=20\times\dfrac{\sqrt{3}}{2}=10\sqrt{3}\) cm\(^{2}\) [A1] cao.

The same pair of sides and the same angle serve both parts, but through different ratios: cosine for the length, sine for the area. Mixing them up is the usual error.

Because \(\cos 60^{\circ}\) is exactly \(\dfrac{1}{2}\), the cosine rule reduces here to \(BC^{2}=b^{2}+c^{2}-bc\), which is why the answer comes out as a whole number without a calculator.