Use the sine rule to find unknown sides and angles in non-right-angled triangles, including the ambiguous case

Use the sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B}$.

$\frac{8}{\sin 35\degree} = \frac{b}{\sin 70\degree}$.

$b = \frac{8 \sin 70\degree}{\sin 35\degree} = \frac{8 \times 0.9397}{0.5736} \approx \frac{7.518}{0.5736} \approx 13.105$ cm.

Rounded to the nearest mm: $b \approx 13.1$ cm.

Markers reward correct statement of the rule, substitution of given values, and an answer rounded to the requested precision.

Use $\frac{\sin Y}{y} = \frac{\sin X}{x}$.

$\sin Y = \frac{9 \sin 110\degree}{14} = \frac{9 \times 0.9397}{14} = \frac{8.4573}{14} \approx 0.6041$.

$Y = \sin^{-1}(0.6041) \approx 37.2\degree$.

The ambiguous case does not arise because angle $X = 110\degree$ is obtuse, so the other two angles must both be acute. $Y \approx 37.2\degree$ is the only valid solution.

Markers reward the sine rule applied correctly, the angle, and the justification that an obtuse angle in the triangle forces the other angles to be acute.

Choose the matched pairs. You know the side $a$ with its opposite angle $A$, and you want the side $b$ opposite $B$. That is two complete pairs, so the side-on-top form of the sine rule links them:

Make $b$ the subject and substitute. Multiply both sides by $\sin 65\degree$:

Check. Angle $B = 65\degree$ is larger than $A = 40\degree$, so the side opposite it should be the longer one, and $14.1 > 10$. Correct.

Choose the matched pairs. You know the side $c$ with its opposite angle $C$, and you want the side $a$ opposite $A$. That is two complete pairs, so the side-on-top form of the sine rule links them directly:

Make $a$ the subject and substitute. Multiply both sides by $\sin 55\degree$:

Check. Angle $A = 55\degree$ is larger than $C = 48\degree$, so side $a$ should be longer than $c = 9$, and $9.9 > 9$. Correct.

Answer: $a \approx 9.9$ cm.

Set up the sine-on-top form. The unknown is an angle, so put the sines on top. Pair side $a$ with angle $A$ and side $c$ with angle $C$:

Evaluate

$\sin A = \frac{11 \times 0.9659}{18} = \frac{10.625}{18} \approx 0.5903$, so $A = \sin^{-1}(0.5903) \approx 36.2\degree$.

No ambiguity here

The given angle $C = 105\degree$ is obtuse, so the other two angles must both be acute. The obtuse alternative $180\degree - 36.2\degree = 143.8\degree$ would make $A + C$ far more than $180\degree$, so it is impossible.

Check

Side $a = 11$ is shorter than $c = 18$, so angle $A$ should be smaller than $C = 105\degree$, and $36.2\degree$ is. Correct.

Answer

$A \approx 36.2\degree$.

Set up the sine-on-top form. The unknown is an angle, so put the sines on top. Pair side $b$ with angle $B$ and side $a$ with angle $A$:

Evaluate

$\sin B = \frac{14 \times 0.6157}{20} = \frac{8.6202}{20} \approx 0.4310$, so $B = \sin^{-1}(0.4310) \approx 25.5\degree$.

Test the obtuse alternative

The other angle with this sine is $180\degree - 25.5\degree = 154.5\degree$, but $A + 154.5\degree = 38\degree + 154.5\degree = 192.5\degree > 180\degree$, which is impossible. So the obtuse case is rejected and only one triangle exists. (This matches the rule that when the side opposite the known angle, $a = 20$, is longer than the other given side, $b = 14$, the case is not ambiguous.)

Check

Side $b = 14$ is shorter than $a = 20$, so angle $B \approx 25.5\degree$ should be smaller than $A = 38\degree$, and it is. Correct.

Find the third angle first. The wanted side $c$ is opposite $C$, but $C$ is not given. Use the angle sum:

Set up the matched pairs. Now you have side $a = 24$ opposite $A = 58\degree$, and you want side $c$ opposite $C = 75\degree$. Use the side-on-top form:

Solve for $c$. Multiply both sides by $\sin 75\degree$:

Check. Angle $C = 75\degree$ is the largest angle, so $c$ should be the longest side, and $27.3$ is longer than $a = 24$. Correct.

Answer: $c \approx 27.3$ cm.

Identify the matched pairs. The side $a = 15$ sits opposite the known angle $A = 52\degree$, and the wanted side $b$ sits opposite $B = 43\degree$. Both are complete pairs, so the sine rule applies directly without needing the third angle:

Solve for $b$. Multiply both sides by $\sin 43\degree$:

Check. Angle $B = 43\degree$ is smaller than $A = 52\degree$, so $b$ should be shorter than $a = 15$, and $12.98 < 15$. Correct.

Find the angle at $P$. The angle inside the triangle at $P$ is the difference of the two bearings measured from $P$:

Find the angle at $L$ using a back-bearing. The bearing of $P$ seen from $L$ is the back-bearing of $035\degree$, that is $035\degree + 180\degree = 215\degree$. The bearing from $L$ to $M$ is $150\degree$, so the interior angle at $L$ is:

The third angle is $\angle M = 180\degree - 75\degree - 65\degree = 40\degree$.

Apply the sine rule. The wanted side $PM$ is opposite $\angle L$, and the known side $PL = 8$ km is opposite $\angle M$:

Check. $\angle L = 65\degree$ is larger than $\angle M = 40\degree$, so $PM$ should be longer than $PL = 8$, and $11.28 > 8$. Correct.

Set up the sine-on-top form. The unknown is angle $B$, so put the sines on top, pairing $b$ with $B$ and $a$ with $A$:

Write down both candidate angles. Since $0 < 0.7272 < 1$, two angles between $0\degree$ and $180\degree$ have this sine:

Test each against the angle sum

For $B_1$: $A + B_1 = 29\degree + 46.7\degree = 75.7\degree < 180\degree$, leaving $C \approx 104.3\degree$. Valid. For $B_2$: $A + B_2 = 29\degree + 133.3\degree = 162.3\degree < 180\degree$, leaving $C \approx 17.7\degree$. Also valid. Both survive, so this is the ambiguous case (it arises because $b = 15$ is longer than $a = 10$ with $A$ acute).

Check

Both values of $B$ have a sine of about $0.7272$ on a calculator, and both leave a positive third angle. Correct.

Answer

$B \approx 46.7\degree$ or $B \approx 133.3\degree$.

Part (a), find the angle at $C$. The baseline $AB$ is the side opposite $C$. Use the angle sum of the triangle:

Find $BC$ with the sine rule. The distance $BC$ is the side opposite $A = 62\degree$, and the known baseline $AB = 140$ m is opposite $C = 69\degree$:

Part (b), find $AC$. The distance $AC$ is the side opposite $B = 49\degree$, with the same known side $AB = 140$ m opposite $C = 69\degree$:

Check. Angle $C = 69\degree$ is the largest angle, so the baseline $AB = 140$ opposite it should be the longest side, and both $BC \approx 132.4$ and $AC \approx 113.2$ are shorter. Correct.

Answer: (a) angle $A\hat{C}B = 69\degree$ and $BC \approx 132.4$ m; (b) $AC \approx 113.2$ m.

Find the angle at $S$. The interior angle at $S$ is the difference of the two bearings taken from $S$ (to $A$ and to $B$):

Find the angle at $A$ with a back-bearing. The bearing of $S$ seen from $A$ is the back-bearing of $070\degree$, that is $070\degree + 180\degree = 250\degree$. The bearing from $A$ to $B$ is $130\degree$, so the interior angle at $A$ is:

The third angle is $\angle B = 180\degree - 25\degree - 120\degree = 35\degree$.

Apply the sine rule. The wanted side $SB$ is opposite $\angle A$, and the known side $SA = 12$ km is opposite $\angle B$:

Check. $\angle A = 120\degree$ is the largest angle, so $SB$ (opposite it) must be the longest side, and $18.12$ is greater than $SA = 12$. Correct.

Find angle $A$ with the sine-on-top form. Pair $a$ with $A$ and $b$ with $B$:

So $A = \sin^{-1}(0.6931) \approx 43.9\degree$.

Explain why there is no ambiguity. The other angle with this sine is $180\degree - 43.9\degree = 136.1\degree$, but $B + 136.1\degree = 104\degree + 136.1\degree = 240.1\degree > 180\degree$, which is impossible. Because the given angle $B = 104\degree$ is obtuse, the other two angles must both be acute, so only $A \approx 43.9\degree$ is valid.

Find the third angle, then side $c$. $C = 180\degree - 104\degree - 43.9\degree = 32.1\degree$. Now use a matched pair to get $c$ (opposite $C$), using $b$ opposite $B$:

Check. $C = 32.1\degree$ is the smallest angle, so $c \approx 11.51$ should be the shortest side, and it is shorter than both $a = 15$ and $b = 21$. Correct.

Part (a), set up the sine-on-top form. The unknown is angle $B$, so pair $b$ with $B$ and $a$ with $A$:

Find both candidate angles. Since $0 < 0.6028 < 1$:

Keep the ones that fit the angle sum. For $B_1$: $A + B_1 = 26\degree + 37.1\degree = 63.1\degree < 180\degree$. Valid. For $B_2$: $A + B_2 = 26\degree + 142.9\degree = 168.9\degree < 180\degree$. Also valid. Both survive (ambiguous, since $b = 11 > a = 8$ with $A$ acute), so $B \approx 37.1\degree$ or $B \approx 142.9\degree$.

Part (b), find $c$ for each triangle. First the third angle in each case. Case 1: $C_1 = 180\degree - 26\degree - 37.1\degree = 116.9\degree$. Case 2: $C_2 = 180\degree - 26\degree - 142.9\degree = 11.1\degree$. Then use the pair $a$ with $A$:

Case 1: $c_1 = \frac{8 \sin 116.9\degree}{\sin 26\degree} = \frac{8 \times 0.8918}{0.4384} \approx 16.27 \text{ km}.$

Case 2: $c_2 = \frac{8 \sin 11.1\degree}{\sin 26\degree} = \frac{8 \times 0.1925}{0.4384} \approx 3.51 \text{ km}.$

Check. In Case 1 the largest angle is $C_1 = 116.9\degree$, so $c_1 \approx 16.27$ should be the longest side, and it is. In Case 2 the smallest angle is $C_2 = 11.1\degree$, so $c_2 \approx 3.51$ should be the shortest side, and it is. Correct.