Find the angle opposite each marked side by looking along the triangle. The side \(16\) is opposite the angle \(51^\circ\), and the side \(x\) is opposite the angle \(62^\circ\).

The sine rule pairs each side with the sine of the angle opposite it, so the two pairs are \(16\) cm with \(\sin 51^\circ\), and \(x\) cm with \(\sin 62^\circ\).

Each pair makes one fraction, both with the sides on top. The equation that sets the two fractions equal is:

\(\frac{ x }{ \sin 62^\circ } = \frac{ 16 }{ \sin 51^\circ }\)

The other equations pair a side with an angle that is not opposite it, leave a sine out, or turn only one fraction upside down.

Pair each side with the angle opposite it. \(AC\) is opposite the angle at \(B\), and \(BC\) is opposite the angle at \(A\), so both pairs are known and the sine rule can be used straight away.

\(\frac{AC}{\sin B} = \frac{BC}{\sin A}\)

\(AC = \frac{ 7 \times \sin 78^\circ }{ \sin 65^\circ }\)

\(AC = 7.55\) cm

\(x\) is the angle at \(B\), and the side opposite it is \(CA = 12\) cm. The angle at \(A\) is known and the side opposite it is \(BC = 13\) cm, so both pairs are complete.

Put the sines on top, because it is an angle that is missing:

\(\frac{\sin x}{ 12 } = \frac{\sin 61^\circ}{ 13 }\)

\(\sin x = \frac{ 12 \times \sin 61^\circ }{ 13 }\)

\(x = \sin^{-1}\left( \frac{ 12 \times \sin 61^\circ }{ 13 } \right) = 53.8^\circ\)

\(x\) is opposite the shorter of the two sides, so it is the smaller angle — the acute answer is the one wanted.

The three angles of a triangle add to \(180^\circ\).

\(180 - 39 - 44 = 97\)

Angle \(ACB = 97^\circ\).

\(AB\) is the only side given, and the angle opposite it is the one found in part (a). That is why the third angle had to come first: without it there is no complete pair to start from.

\(\frac{BC}{\sin A} = \frac{AB}{\sin C}\)

\(BC = \frac{ 8 \times \sin 39^\circ }{ \sin 97^\circ }\)

\(BC = 5.1\) cm

Only one side is known, and the angle opposite it is the angle at \(R\), which is not given. Find that first.

\(\angle PRQ = 180 - 59 - 70 = 51^\circ\)

Now \(PQ\) and the angle at \(R\) make a complete pair, and \(QR\) is opposite the angle at \(P\).

\(\frac{QR}{\sin P} = \frac{PQ}{\sin R}\)

\(QR = \frac{ 860 \times \sin 59^\circ }{ \sin 51^\circ }\)

\(QR = 949\) m, to the nearest metre.

\(QR\) is opposite the angle at \(P\), and \(PQ\) is opposite the angle at \(R\), so those two pairs are what the sine rule joins.

\(\frac{\sin R}{ 22 } = \frac{\sin 78^\circ}{ 25 }\)

\(\sin R = \frac{ 22 \times \sin 78^\circ }{ 25 }\)

The inverse sine offers the acute angle, and here that is the one wanted: \(PQ\) is the shorter of the two sides, so the angle facing it is the smaller of the two and cannot be the obtuse one.

\(\angle PRQ = 59.4^\circ\)

No more trigonometry is needed — the three angles add to \(180^\circ\).

\(180 - 78 - 59.4 = 42.6\)

\(\angle PQR = 42.6^\circ\)

\(AB\) is due north, so the bearing of \(A\) from \(B\) is \(180^\circ\). Angle \(ABC\) is the gap between that and the bearing of \(C\):

\(180 - 40 = 140\), so angle \(ABC = 140^\circ\)

Use the cosine rule in triangle \(ABC\):

\(CA^2 = 47^2 + 70^2 - 2 \times 47 \times 70 \times \cos 140^\circ \approx 12149.57\)

\(CA = \sqrt{ 12149.57 } \approx 110.23\) km

\(CA \approx 110\) km to the nearest km, as stated.

Use the sine rule in triangle \(ABC\) with the unrounded value \(CA \approx 110.23\) km. Angle \(ACB\) is opposite \(AB\), the shorter of the two known sides, so it is acute:

\(\frac{ \sin ACB }{ 47 } = \frac{ \sin 140^\circ }{ 110.23 }\)

\(\sin ACB \approx 0.274\), so angle \(ACB \approx 15.9^\circ\)

The bearing of \(B\) from \(C\) is \(40 + 180 = 220\). The point \(A\) lies anticlockwise from \(B\) as seen from \(C\), by angle \(ACB\):

\(220 - 15.9 \approx 204.1\)

The bearing of \(A\) from \(C\) is \(204^\circ\) to the nearest degree.

\(180 - 59 - 67 = 54\)

Angle \(ACB = 54^\circ\).

The perimeter needs both missing sides, so use the sine rule twice against the same complete pair, \(AB\) with the angle at \(C\).

\(BC = \frac{ 8 \times \sin 59^\circ }{ \sin 54^\circ } = 8.48\) cm, to 2 d.p.

\(CA = \frac{ 8 \times \sin 67^\circ }{ \sin 54^\circ } = 9.10\) cm, to 2 d.p.

Add all three sides. Keep both of those lengths in the calculator at full accuracy and round only the total — adding the two rounded values above can leave the last digit one out.

\(\text{Perimeter} = 25.6\) cm

The sine rule needs a side and the angle opposite it, and there is no such pair here — the only angle known sits between the two known sides. So it has to be the cosine rule.

\(AC^2 = 8^2 + 7^2 - 2 \times 8 \times 7 \times \cos 82^\circ\)

\(AC^2 = 64 + 49 - 112\cos 82^\circ = 97.41\)

\(AC = \sqrt{ 97.41 } = 9.9\) cm

Now that \(AC\) is known there is a complete pair — \(AC\) with the angle at \(B\) — so the sine rule finishes it.

\(\frac{\sin A}{ 7 } = \frac{\sin 82^\circ}{ AC }\)

\(\sin A = \frac{ 7 \times \sin 82^\circ }{ AC }\)

Leave \(AC\) as it stands in the calculator. Typing \(9.9\) back in can move the angle by three degrees, and just occasionally makes \(\sin A\) come out above 1 so the calculator refuses it — which is why the line above names \(AC\) rather than printing it.

\(\angle BAC = 45^\circ\), to the nearest degree.

\(BC\) is opposite the angle at \(A\), and \(AB\) is opposite the angle at \(C\), so the two pairs are complete and the sine rule can start.

\(\frac{\sin C}{ 12 } = \frac{\sin 30^\circ}{ 7 }\)

\(\sin C = \frac{ 12 \times \sin 30^\circ }{ 7 }\)

A calculator gives \(\sin^{-1}\) of that as \(59.0^\circ\), which is acute. That is not the answer here.

Two angles between \(0^\circ\) and \(180^\circ\) share the same sine, and the second one is \(180^\circ\) minus the first. The question says angle \(ACB\) is obtuse, so it is that second one:

\(\angle ACB = 180 - 59.0 = 121.0^\circ\)