In triangle \(PQR\), \(PQ=6\) cm, \(QR=10\) cm and angle \(PQR=120^{\circ}\). (a) Show that \(PR=14\) cm. [3] (b) Find the exact area of triangle \(PQR\). [...

Two sides and the angle between them are given, which is exactly the case for the cosine rule. The angle \(120^{\circ}\) is obtuse, so its cosine is negative, and the exact values \(\cos 120^{\circ}=-\dfrac{1}{2}\) and \(\sin 120^{\circ}=\dfrac{\sqrt{3}}{2}\) make the whole question possible without a calculator.

(a) The side \(PR\) is opposite the given angle \(PQR\), so

\(PR^{2}=6^{2}+10^{2}-2\times 6\times 10\times\cos 120^{\circ}\) [M1]

Using \(\cos 120^{\circ}=-\dfrac{1}{2}\) [M1], the final term becomes \(-120\times\left(-\dfrac{1}{2}\right)=+60\), so

\(PR^{2}=36+100+60=196\), hence \(PR=\sqrt{196}=14\) cm [A1]

The sign is the whole point here: because the angle is obtuse, the subtraction turns into an addition and \(PR\) comes out longer than either given side, as it must be when the angle between them is wide.

(b) Area of a triangle from two sides and the included angle is \(\dfrac{1}{2}ab\sin C\):

\(\dfrac{1}{2}\times 6\times 10\times\dfrac{\sqrt{3}}{2}\) [M1]

\(\dfrac{1}{2}\times 6\times 10=30\), and \(30\times\dfrac{\sqrt{3}}{2}=15\sqrt{3}\).

Area \(=15\sqrt{3}\) cm\(^{2}\) [A1]

"Exact" means the surd must be left in place; a rounded decimal would not earn the mark.

(c) Apply the sine rule, pairing each angle with the side opposite it. Angle \(QPR\) is opposite \(QR=10\), and angle \(PQR=120^{\circ}\) is opposite \(PR=14\):

\(\dfrac{\sin QPR}{10}=\dfrac{\sin 120^{\circ}}{14}\), so \(\sin QPR=\dfrac{10}{14}\times\dfrac{\sqrt{3}}{2}=\dfrac{10\sqrt{3}}{28}=\dfrac{5\sqrt{3}}{14}\)

\(\sin QPR=\dfrac{5\sqrt{3}}{14}\) oe [B1]