Side AC is opposite angle B and the required side AB is opposite angle C. So we write the sine rule as follows:

$$ \begin{eqnarray} \frac{AC}{sin\ B} &=& \frac{AB}{sin\ C} \\[9pt] \frac{26}{sin\ 114^\circ} &=& \frac{AB}{sin\ 32^\circ} \\[9pt] AB\ sin\ 114^\circ &=& 26\ sin\ 32^\circ\:\:\small\textsf{(cross-multiplying)}\normalsize \\[6pt] AB &=& \frac{26\ sin\ 32^\circ}{sin\ 114^\circ} \\[9pt] AB &=& 15.0817... \\[6pt] AB &\approx& 15.1\ \small\textsf{mm} \end{eqnarray} $$

This is a common Paper 1 question type. We are not told the size, in degrees, of the angles Q or R. Instead we are told sin Q and sin R directly.

Side PR is opposite angle Q and side PQ is opposite angle R. So we rewrite the sine rule with these letters:

$$ \begin{eqnarray} \frac{PR}{sin\ Q} &=& \frac{PQ}{sin\ R} \\[9pt] \frac{16}{0.8} &=& \frac{PQ}{0.7} \\[9pt] 0.7 \times 16 &=& 0.8 \times PQ\:\:\small\textsf{(cross-multiplying)}\normalsize \\[6pt] 7 \times 16 &=& 8 \times PQ\:\:\small\textsf{(both sides x10)}\normalsize \\[6pt] PQ &=& \frac{7 \times 16}{8} \\[6pt] PQ &=& 7 \times 2 \:\:\small\textsf{(simplifying the fraction)}\normalsize\\[6pt] PQ &=& 14\ \small\textsf{cm} \\[6pt] \end{eqnarray} $$

There has never been an exam question like this, but the course specification  states explicitly that "related angles could be combined with sine and cosine rules" so you need to be able to handle this.

This question needs an understanding of the so-called "ambiguous case" of the sine rule. There are actually two different ways to draw a triangle with AB = 6 cm, angle A = 12° and BC = 2.5 cm – try it and see! So there are two different possible angles for C: one acute and one obtuse.

However, we'll worry about that at the end. We start in the normal way:

$$ \begin{eqnarray} \frac{AB}{sin\ C} &=& \frac{BC}{sin\ A} \\[9pt] \frac{6}{sin\ C} &=& \frac{2.5}{sin\ 12^\circ} \\[9pt] 2.5\ sin\ C &=& 6\ sin\ 12^\circ\:\:\small\textsf{(cross-multiplying)}\normalsize \\[6pt] sin\ C &=& \frac{6\ sin\ 12^\circ}{2.5}\\[9pt] C &=& sin^{-1}\ \frac{6\ sin\ 12^\circ}{2.5} \\[9pt] C &\approx& 29.9^\circ \\[6pt] \end{eqnarray} $$

So that's one possible value of C. But from our knowledge of the quadrant diagram from studying trig equations, there is a related angle in the 2nd quadrant. It is 180 – 29.9 ° = 150.1°.

Therefore, angle C = 29.9° or 150.1° (correct to 1 decimal place).

A question very similar to this appeared in 2019 Paper 2. It caught a lot of candidates out, because they weren't expecting to have to use SOH CAH TOA (a National 4 topic) as well as the sine rule.

The method here is to find either FD or SD, then draw the vertical height below D and use right-angled trigonometry to find the height.

We will find FD. You might like to try this yourself, but finding SD instead.

First, we need to find angle D, because it's opposite the 54 metres:

$$ D=180-38-49=93^\circ $$

Now we're ready to use the sine rule:

$$ \begin{eqnarray} \frac{FD}{sin\ S^\circ} &=& \frac{FS}{sin\ D^\circ} \\[9pt] \frac{FD}{sin\ 49^\circ} &=& \frac{54}{sin\ 93^\circ} \\[9pt] FD\ sin\ 93^\circ &=& 54\ sin\ 49^\circ\:\:\small\textsf{(cross-multiplying)}\normalsize \\[6pt] FD &=& \frac{54\ sin\ 49^\circ}{sin\ 93^\circ} \\[9pt] FD &=& 40.8102...\ \small\textsf{metres} \end{eqnarray} $$

Don't round this yet! Keep all the accuracy in your calculator to avoid a rounding error.

Finally, we drop a vertical line from D and use right-angled trigonometry (SOH CAH TOA) to find its height h.

$$ \begin{eqnarray} sin\ 38^\circ &=& \frac{h}{40.8102} \\[6pt] h &=& 40.8102\ sin\ 38^\circ \\[6pt] &=& 25.125... \\[6pt] &\approx& 25.1\ \small\textsf{metres} \end{eqnarray} $$

Note that are not told the size, in degrees, of the angles K or L.

Instead, we are told the values of sin K and sin L, so that we don't need a calculator.

$$ \begin{eqnarray} \frac{KM}{sin\ L} &=& \frac{LM}{sin\ K} \\[9pt] \frac{18}{0.9} &=& \frac{LM}{0.4} \\[9pt] 0.4 \times 18 &=& 0.9 \times LM\:\:\small\textsf{(cross-multiplying)}\normalsize \\[6pt] 4 \times 18 &=& 9 \times LM\:\:\small\textsf{(both sides x10)}\normalsize \\[6pt] LM &=& \frac{4 \times 18}{9} \\[9pt] LM &=& 4 \times 2 \:\:\small\textsf{(simplifying the fraction)}\normalsize\\[6pt] LM &=& 8\ \small\textsf{cm} \\[6pt] \end{eqnarray} $$

This 3-mark question needs us to calculate the angle CBD before using the sine rule.

The question really should have told us that ABC is a straight line, but as answering it would be impossible without knowing this, we can safely assume it. So \(\angle\)CBD \(=180-75=105^\circ\small.\)

Now we can use the sine rule:

$$ \begin{eqnarray} \frac{DC}{sin\ B} &=& \frac{BC}{sin\ D} \\[9pt] \frac{DC}{sin\ 105^\circ} &=& \frac{20}{sin\ 37^\circ} \\[9pt] DC\ sin\ 37^\circ &=& 20\ sin\ 105^\circ \\[6pt] DC &=& \frac{20\ sin\ 105^\circ}{sin\ 37^\circ} \\[9pt] DC &=& 32.10042... \\[6pt] DC &\approx& 32.1\ \small\textsf{cm} \end{eqnarray} $$

First, we need to use the sine rule to find the length of either KB or BM. It doesn't matter which. We will find KB.

Note that angle KBM = 180 – 52 – 34 = 94°. We need this because it is opposite the given side.

Now we're ready to use the sine rule:

$$ \begin{eqnarray} \frac{KB}{sin\ M^\circ} &=& \frac{KM}{sin\ B^\circ} \\[9pt] \frac{KB}{sin\ 34^\circ} &=& \frac{350}{sin\ 94^\circ} \\[9pt] KB\ sin\ 94^\circ &=& 350\ sin\ 34^\circ\:\:\small\textsf{(cross-multiplying)}\normalsize \\[6pt] KB &=& \frac{350\ sin\ 34^\circ}{sin\ 94^\circ} \\[9pt] KB &=& 196.195...\ \small\textsf{metres} \end{eqnarray} $$

Don't round this yet! Keep all the accuracy in your calculator to avoid a rounding error.

Finally, we drop a vertical line from B and use right-angled trigonometry (SOH CAH TOA) to find its height h.

$$ \begin{eqnarray} sin\ 52^\circ &=& \frac{h}{196.195} \\[6pt] h &=& 196.195\ sin\ 52^\circ \\[6pt] &=& 154.604... \\[6pt] &\approx& 154.6\ \small\textsf{metres} \end{eqnarray} $$

This 3-mark question is a simple application of the sine rule to find an angle.

$$ \begin{eqnarray} \frac{QR}{sin\ P} &=& \frac{PR}{sin\ Q} \\[9pt] \frac{9.8}{sin\ 54^\circ} &=& \frac{11.3}{sin\ Q^\circ} \\[9pt] 9.8\ sin\ Q^\circ &=& 11.3\ sin\ 54^\circ \\[6pt] sin\ Q &=& \frac{11.3\ sin\ 54^\circ}{9.8}\\[9pt] Q &=& sin^{-1}\ \frac{11.3\ sin\ 54^\circ}{9.8} \\[9pt] Q &\approx& 68.9^\circ \\[6pt] \end{eqnarray} $$

This 3-mark question is very similar to the 2021 question above.

$$ \begin{eqnarray} \frac{KL}{sin\ J} &=& \frac{JL}{sin\ K} \\[9pt] \frac{7}{sin\ 25^\circ} &=& \frac{10}{sin\ K^\circ} \\[9pt] 7\ sin\ K^\circ &=& 10\ sin\ 25^\circ \\[6pt] sin\ K &=& \frac{10\ sin\ 25^\circ}{7}\\[9pt] K &=& sin^{-1}\ \frac{10\ sin\ 25^\circ}{7} \\[9pt] K &\approx& 37.1^\circ \\[6pt] \end{eqnarray} $$