The angle that is given lies between the two sides that are given, and the side wanted is opposite it. That is exactly the cosine rule.

\(a^2 = b^2 + c^2 - 2bc\cos A\)

\(BC^2 = 15^2 + 10^2 - 2 \times 15 \times 10 \times \cos 68^\circ\)

\(BC^2 = 225 + 100 - 300\cos 68^\circ = 212.62\)

\(BC = \sqrt{ 212.62 } = 14.6\)

\(BC = 14.6\) cm

The sine rule can only be started from a side together with the angle opposite it. Here the only angle known is at \(B\), and the side opposite \(B\) is \(AC\) — which is the unknown. So there is no complete pair and the sine rule cannot begin.

Two sides and the angle between them is the cosine rule's own case, so use the cosine rule.

\(AC^2 = 9^2 + 16^2 - 2 \times 9 \times 16 \times \cos 127^\circ\)

\(AC^2 = 81 + 256 - 288\cos 127^\circ = 510.32\)

\(AC = \sqrt{ 510.32 } = 22.6\)

\(AC = 22.6\) cm

All three sides are known and an angle is wanted, so use the rearranged cosine rule. Take \(a\) to be the side opposite \(x\), which is \(BC = 6\) cm.

\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)

\(\cos x = \frac{ 49 + 81 - 36 }{ 126 } = \frac{ 94 }{ 126 }\)

\(x = \cos^{-1}\left( \frac{ 94 }{ 126 } \right) = 41.8^\circ\)

If \(\cos x\) comes out negative that is not a mistake — it simply means the angle is obtuse.

Sketch the triangle. The two paths are two sides of it, the angle between them is known, and the distance wanted is the third side — so the cosine rule does it in one step.

\(d^2 = 11^2 + 13^2 - 2 \times 11 \times 13 \times \cos 101^\circ\)

\(d^2 = 121 + 169 - 286\cos 101^\circ = 344.57\)

\(d = \sqrt{ 344.57 } = 18.6\)

They are \(18.6\) km apart.

The largest angle in any triangle is the one opposite the longest side. The longest side here is \(BC = 55\) m, so the largest angle is \(\angle BAC\).

\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)

\(\cos A = \frac{ 2025 + 900 - 3025 }{ 2700 } = \frac{ -100 }{ 2700 }\)

\(A = \cos^{-1}\left( \frac{ -100 }{ 2700 } \right) = 92.1^\circ\)

The largest angle is \(92.1^\circ\).

\(BD\) is a side of triangle \(ABD\), so that is the triangle to work in. Its two other sides are \(AB\) and \(AD\), and \(AD = BC = 13\) cm because opposite sides of a parallelogram are equal.

The angle between those two sides is the angle at \(A\), which is not the angle given. \(AD\) and \(BC\) are parallel, so the angles at \(A\) and at \(B\) are co-interior and add to \(180^\circ\):

\(\angle DAB = 180 - 124 = 56^\circ\)

Now the cosine rule applies to triangle \(ABD\):

\(BD^2 = 17^2 + 13^2 - 2 \times 17 \times 13 \times \cos 56^\circ\)

\(BD^2 = 289 + 169 - 442\cos 56^\circ = 210.84\)

\(BD = \sqrt{ 210.84 } = 14.5\) cm

Using \(124^\circ\) instead would give the other diagonal, \(AC\).

The north lines at \(A\) are all parallel, and both bearings are measured clockwise from north at \(A\), so the angle between \(AB\) and the fishing boat's course is the difference of the bearings:

\(284 - 217 = 67\)

Distance sailed: \(2\) hours \(30\) minutes is \(2.5\) hours, so

\(11 \times 2.5 = 27.5\) miles

Let \(x\) be the fishing boat's distance from \(B\). Use the cosine rule in the triangle formed by \(A\), \(B\) and the fishing boat's position, with the angle \(67^\circ\) at \(A\):

\(x^2 = 47^2 + 27.5^2 - 2 \times 47 \times 27.5 \times \cos 67^\circ\)

\(x^2 \approx 1955.2\)

\(x = \sqrt{ 1955.2 } \approx 44.2\) miles

The question is whether the fishing boat is closer to \(B\) than when it left \(A\), when it was \(47\) miles away. The distance sailed is not the quantity to compare. Compare \(44.2\) miles with \(47\) miles: \(44.2\) is less than \(47\).

Yes.

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.

\(AC^2 = 12^2 + 15^2 - 2 \times 12 \times 15 \times \cos 88^\circ\)

\(AC^2 = 144 + 225 - 360\cos 88^\circ = 356.44\)

\(AC = \sqrt{ 356.44 } = 18.9\) cm

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

\(\frac{\sin A}{ 15 } = \frac{\sin 88^\circ}{ AC }\)

\(\sin A = \frac{ 15 \times \sin 88^\circ }{ AC }\)

Leave \(AC\) as it stands in the calculator. Typing \(18.9\) 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.

\(\angle BAC = 53^\circ\), to the nearest degree.

The cosine rule needs the angle between the two known sides, which here is angle \(ABC\). Neither bearing is that angle, so find it first.

The bearing of \(A\) from \(B\) is the back bearing of the first leg:

\((127 + 180)^\circ = 307^\circ\)

The two north lines are parallel, which is what makes that rule work. Angle \(ABC\) is the turn from \(BC\) round to \(BA\):

\(\angle ABC = 307 - 186 = 121^\circ\)

Now the two sides that meet at \(B\) are known and the side wanted is opposite that angle.

\(AC^2 = 22^2 + 25^2 - 2 \times 22 \times 25 \times \cos 121^\circ\)

\(AC^2 = 484 + 625 - 1100\cos 121^\circ = 1675.54\)

\(AC = \sqrt{ 1675.54 } = 40.9\) km

\(A\), \(B\) and \(C\) are three corners in a row, so \(AC\) is the third side of triangle \(ABC\). Two of its sides are sides of the nonagon, \(AB = BC = 11\) cm, and the angle between them is the interior angle of the nonagon at \(B\).

The exterior angles of any polygon add to \(360^\circ\), and a regular nonagon has 9 equal ones:

\(\text{exterior angle} = \frac{360}{ 9 } = 40^\circ\)

An interior angle and its exterior angle lie on a straight line:

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

Two sides and the angle between them, with the third side wanted, is the cosine rule:

\(AC^2 = 11^2 + 11^2 - 2 \times 11 \times 11 \times \cos 140^\circ\)

\(AC^2 = 121 + 121 - 242\cos 140^\circ = 427.38\)

\(AC = \sqrt{ 427.38 } = 20.7\) cm