5.1 The sine and cosine rules and the area of a triangle — Practise

Work through these on your own. Decide which rule you need before you reach for the calculator: two angles and a side, two sides and the included angle, or three sides. Give each angle in degrees and each length to three significant figures unless the question says otherwise.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

In triangle $ABC$ , angle $A = 30^\circ$ , angle $B = 45^\circ$ and side $a = 6$ cm. Find side $b$ .
In triangle $ABC$ , side $a = 5$ cm, side $b = 5\sqrt{2}$ cm and angle $A = 30^\circ$ . Find the acute value of angle $B$ .
In triangle $ABC$ , side $b = 5$ cm, side $c = 8$ cm and the included angle $A = 60^\circ$ . Find side $a$ .
A triangle has sides $3$ cm, $5$ cm and $7$ cm. Find the angle opposite the $7$ cm side, in degrees.
Find the area of the triangle with $a = 6$ cm, $b = 8$ cm and included angle $C = 30^\circ$ .
Find the area of the triangle with sides $6$ cm and $4\sqrt{2}$ cm and included angle $45^\circ$ .
In triangle $ABC$ , side $a = 10$ cm, side $b = 12$ cm and angle $A = 40^\circ$ . Find the obtuse possible value of angle $B$ , to three significant figures.
In triangle $SLB$ , the angle at $S$ is $30^\circ$ , the angle at $B$ is $45^\circ$ and the side $SL = 10$ km. Find the side $LB$ , opposite the angle at $S$ .
In triangle $ABC$ , side $b = 7$ cm, side $c = 9$ cm and the included angle $A = 120^\circ$ . Find side $a$ , in surd form.
Find the area of the triangle with sides $5$ cm and $8$ cm and included angle $150^\circ$ .

Answers and concise solutions

1. The side $b$

Answer: Pair each side with the angle opposite it: $\dfrac{b}{\sin 45^\circ} = \dfrac{6}{\sin 30^\circ}$ . So $b = \dfrac{6 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 6\sqrt{2} \approx 8.49$ cm.

2. The acute angle $B$

Answer: Use the angle form: $\dfrac{\sin B}{5\sqrt{2}} = \dfrac{\sin 30^\circ}{5}$ . Then $\sin B = \dfrac{5\sqrt{2} \times \frac{1}{2}}{5} = \dfrac{\sqrt{2}}{2}$ . The acute value is $B = 45^\circ$ .

3. The side $a$

Answer: Cosine rule: $a^2 = 5^2 + 8^2 - 2 \times 5 \times 8 \times \cos 60^\circ = 25 + 64 - 40 = 49$ , so $a = 7$ cm.

4. The angle opposite the $7$ cm side

Answer: Rearrange the cosine rule: $\cos C = \dfrac{3^2 + 5^2 - 7^2}{2 \times 3 \times 5} = \dfrac{9 + 25 - 49}{30} = -\dfrac{1}{2}$ . So $C = 120^\circ$ , which is $\dfrac{2\pi}{3}$ radians.

5. The area

Answer: Area $= \tfrac{1}{2}ab\sin C = \tfrac{1}{2} \times 6 \times 8 \times \sin 30^\circ = 24 \times \tfrac{1}{2} = 12$ cm $^2$ .

6. The area

Answer: Area $= \tfrac{1}{2} \times 6 \times 4\sqrt{2} \times \sin 45^\circ = 12\sqrt{2} \times \dfrac{\sqrt{2}}{2} = 12$ cm $^2$ .

7. The obtuse angle $B$

Answer: The sine rule gives $\sin B = \dfrac{12\sin 40^\circ}{10} \approx 0.7713$ . The calculator returns the acute value $B \approx 50.5^\circ$ . The other angle with the same sine is $180^\circ - 50.5^\circ \approx 129.5^\circ$ . Both fit here, because the remaining angle is positive in each case.

8. The side $LB$

Answer: Pair $LB$ with the angle at $S$ and $SL$ with the angle at $B$ : $\dfrac{LB}{\sin 30^\circ} = \dfrac{10}{\sin 45^\circ}$ . So $LB = \dfrac{10 \times \frac{1}{2}}{\frac{\sqrt{2}}{2}} = 5\sqrt{2} \approx 7.07$ km.

9. The side $a$

Answer: Cosine rule: $a^2 = 49 + 81 - 126\cos 120^\circ = 130 - 126 \times \left(-\tfrac{1}{2}\right) = 130 + 63 = 193$ . So $a = \sqrt{193} \approx 13.9$ cm. The negative cosine makes the third side longer than either known side.

10. The area

Answer: Area $= \tfrac{1}{2} \times 5 \times 8 \times \sin 150^\circ = 20 \times \tfrac{1}{2} = 10$ cm $^2$ . The angle is obtuse, so its sine is smaller than its acute partner, and the area is smaller than for $30^\circ$ .