Sine Rule, Cosine Rule and Area of a Triangle

You know two sides and the angle between them, so use the cosine rule.

1. QR is opposite the 38^\circ angle
2. Cosine rule: QR^2=7.4^2+10.2^2-2\times 7.4\times 10.2\times\cos 38^\circ
3. QR^2=54.76+104.04-118.95\ldots=39.84\ldots
4. QR=\sqrt{39.84\ldots}=6.312\ldots
5. Answer: 6.31 cm

Pair each side with the angle opposite it, then use the sine rule.

1. BC is opposite angle A (52^\circ) and AC is opposite angle B (71^\circ)
2. Sine rule: \dfrac{AC}{\sin 71^\circ}=\dfrac{9.6}{\sin 52^\circ}
3. AC=\dfrac{9.6\sin 71^\circ}{\sin 52^\circ}=11.518\ldots
4. Answer: 11.5 cm

Find AC first and keep its full calculator value; then split the quadrilateral into two triangles along AC.

Part (a)

1. In triangle ABC you know two sides and the angle between them: use the cosine rule
2. AC^2=8^2+11^2-2\times 8\times 11\times\cos 110^\circ
3. \cos 110^\circ is negative, so AC^2=64+121+60.19\ldots=245.19\ldots
4. AC=15.658\ldots
5. Answer: 15.7 cm

Part (b)

1. Area of triangle ABC=\frac12\times 8\times 11\times\sin 110^\circ=41.346\ldots
2. In triangle ACD: angle ADC=180^\circ-40^\circ-65^\circ=75^\circ
3. Sine rule: \dfrac{AD}{\sin 65^\circ}=\dfrac{AC}{\sin 75^\circ}, so AD=\dfrac{15.658\ldots\times\sin 65^\circ}{\sin 75^\circ}=14.692\ldots
4. Area of triangle ACD=\frac12\times 15.658\ldots\times 14.692\ldots\times\sin 40^\circ=73.940\ldots
5. Total area =41.346\ldots+73.940\ldots=115.28\ldots
6. Answer: 115 cm^2