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

1. Pythagoras: $13^2 - 5^2 = 169 - 25 = 144$, so the third side is $12$ cm.
2. The side opposite $\theta$ is $12$ cm, so $\sin\theta = \dfrac{12}{13}$.
3. Divide by $4$: $x = \dfrac{10\sin 40^\circ}{4} \approx 1.61$.

If question 3 was hard, revise rearranging a formula before you continue.

Two angles and a side are given. That is a job for the sine rule, with $b$ paired with $B$ and $a$ paired with $A$.

Write $\dfrac{b}{\sin 70^\circ} = \dfrac{11}{\sin 50^\circ}$, then multiply both sides by $\sin 70^\circ$.

The side opposite the larger angle is longer, and $13.5 > 11$, which agrees.

The angle sits between the two sides you know. That is the cosine rule, finding the third side.

Substitute into $a^2 = b^2 + c^2 - 2bc\cos A$ with $b = 8$, $c = 11$ and $A = 62^\circ$.

All three sides are known. Use the cosine rule, rearranged to give the angle opposite side $c$.

With $C$ opposite $c = 10$, write $\cos C = \dfrac{6^2 + 7^2 - 10^2}{2 \times 6 \times 7}$.

The angle is obtuse, which is why $\cos C$ is negative.

Use area $= \tfrac{1}{2}bc\sin A$. The angle you need is the one between the two sides.

Substitute the two sides and the angle: $\tfrac{1}{2} \times 9 \times 14 \times \sin 48^\circ$.

This is the sine rule for an angle, so the ambiguous case applies. The calculator will give one acute answer.

Write $\dfrac{\sin Q}{12} = \dfrac{\sin 40^\circ}{9}$, giving $\sin Q \approx 0.8570$. The acute answer is about $59.0^\circ$. The other is $180^\circ - 59.0^\circ = 121.0^\circ$.

The two candidates are $Q \approx 59.0^\circ$ and $Q \approx 121.0^\circ$. Both fit here: with $P = 40^\circ$, the remaining angles are $81.0^\circ$ and $19.0^\circ$, and both are positive. So both triangles are possible.

Draw the triangle $ATP$ and label the two known sides and the angle between them. That is the cosine rule.

$TP^2 = 80^2 + 110^2 - 2 \times 80 \times 110 \times \cos 65^\circ$.