The Sine Rule

The side opposite an angle does not touch the vertex of that angle.

1. The side opposite an angle is the side that does not touch that angle's vertex.
2. \angle ABC is at B. The only side that does not touch B is AC, so AC pairs with \angle ABC.
3. In each other option the side touches the angle's vertex (for example AB meets \angle BAC at A), so it is not an opposite pair.

Match each side to the angle opposite it.

1. Pair each side with the angle opposite it: BC=8 is opposite \angle A=44^\circ, and AC is opposite \angle B=71^\circ.
2. Sine rule: \dfrac{AC}{\sin71^\circ}=\dfrac{8}{\sin44^\circ}.
3. Multiply both sides by \sin71^\circ: AC=\dfrac{8\sin71^\circ}{\sin44^\circ}=10.889\ldots
4. AC=10.9 cm to 1 decimal place.

Match each side to the angle opposite it.

1. AB=12 is opposite \angle C=48^\circ, and BC is opposite \angle A=61^\circ.
2. Sine rule: \dfrac{BC}{\sin61^\circ}=\dfrac{12}{\sin48^\circ}.
3. Multiply both sides by \sin61^\circ: BC=\dfrac{12\sin61^\circ}{\sin48^\circ}=14.123\ldots
4. BC=14.1 cm to 1 decimal place.

Put the unknown angle in the numerator of the sine rule.

1. BC=11 is opposite the known angle \angle BAC=38^\circ, and AC=7 is opposite \angle ABC.
2. To find an angle, write the sine rule with the sines on top: \dfrac{\sin B}{7}=\dfrac{\sin38^\circ}{11}.
3. Multiply by 7: \sin B=\dfrac{7\sin38^\circ}{11}=0.39178\ldots
4. \angle ABC=\sin^{-1}(0.39178\ldots)=23.065\ldots^\circ, which is 23.1^\circ to 1 decimal place.

Write a side above the sine of its opposite angle.

1. AC does not touch B, so AC is opposite \angle ABC=54^\circ.
2. AB=10 does not touch C, so AB is opposite \angle ACB=72^\circ.
3. The sine rule puts each side over the sine of its opposite angle: \dfrac{AC}{\sin54^\circ}=\dfrac{10}{\sin72^\circ}.
4. The second option pairs each side with the wrong angle, and the sine rule uses sine, not cosine.

Match each side to the angle opposite it.

1. BC=16 is opposite \angle A=112^\circ, and AC is opposite \angle B=27^\circ.
2. Sine rule: \dfrac{AC}{\sin27^\circ}=\dfrac{16}{\sin112^\circ}. The sine rule works with obtuse angles too.
3. So AC=\dfrac{16\sin27^\circ}{\sin112^\circ}=7.834\ldots
4. AC=7.8 cm to 1 decimal place.

An obtuse angle is already known, so the required angle must be acute.

1. AC=12 is opposite the known angle \angle ABC=103^\circ, and AB=9 is opposite \angle ACB.
2. Sine rule: \dfrac{\sin C}{9}=\dfrac{\sin103^\circ}{12}, so \sin C=\dfrac{9\sin103^\circ}{12}=0.73077\ldots
3. \sin^{-1}(0.73077\ldots)=46.951\ldots^\circ.
4. The obtuse angle with the same sine, 180^\circ-46.95^\circ=133.05^\circ, is impossible: the triangle already has the obtuse angle 103^\circ. So \angle ACB=47.0^\circ to 1 decimal place.

Find the angle opposite the known side first.

1. The known side AB=6.8 is opposite \angle C, so find \angle C first.
2. Angles in a triangle add to 180^\circ: \angle C=180^\circ-48^\circ-73^\circ=59^\circ.
3. BC is opposite \angle A=48^\circ. Sine rule: \dfrac{BC}{\sin48^\circ}=\dfrac{6.8}{\sin59^\circ}.
4. So BC=\dfrac{6.8\sin48^\circ}{\sin59^\circ}=5.895\ldots, which is 5.9 cm to 1 decimal place.

Multiply both sides by the sine paired with the unknown side.

1. Multiply both sides of \dfrac{a}{\sin A}=\dfrac{b}{\sin B} by \sin B.
2. This gives b=\dfrac{a\sin B}{\sin A}: the unknown side is found by multiplying by the sine of its own opposite angle.
3. The first option has the two sines the wrong way round.

Inverse sine on a calculator gives an acute angle, but the obtuse angle 180^\circ-x has the same sine.

1. BC=8 is opposite the known angle \angle BAC=33^\circ, and AC=11 is opposite \angle ABC.
2. Sine rule: \dfrac{\sin B}{11}=\dfrac{\sin33^\circ}{8}, so \sin B=\dfrac{11\sin33^\circ}{8}=0.74887\ldots
3. \sin^{-1}(0.74887\ldots)=48.493\ldots^\circ. This is acute, but angle ABC is obtuse.
4. Since \sin(180^\circ-x)=\sin x, the obtuse angle with the same sine is 180^\circ-48.493\ldots^\circ=131.506\ldots^\circ.
5. Check: 33^\circ+131.5^\circ is less than 180^\circ, so this triangle exists. \angle ABC=131.5^\circ to 1 decimal place.

Find the angle at the tree before using the sine rule.

1. The known side PQ=80 m is opposite the angle at the tree, R, so find that angle first.
2. \angle PRQ=180^\circ-54^\circ-67^\circ=59^\circ.
3. PR is opposite \angle PQR=67^\circ. Sine rule: \dfrac{PR}{\sin67^\circ}=\dfrac{80}{\sin59^\circ}.
4. So PR=\dfrac{80\sin67^\circ}{\sin59^\circ}=85.911\ldots, which is 85.9 m to 1 decimal place.

Use the sine rule twice, keeping the unrounded side lengths.

1. The known side AB=5.4 is opposite \angle C=180^\circ-53^\circ-76^\circ=51^\circ.
2. BC is opposite \angle A=53^\circ: BC=\dfrac{5.4\sin53^\circ}{\sin51^\circ}=5.5493\ldots m.
3. AC is opposite \angle B=76^\circ: AC=\dfrac{5.4\sin76^\circ}{\sin51^\circ}=6.7421\ldots m.
4. Perimeter =5.4+5.5493\ldots+6.7421\ldots=17.691\ldots m. Add the unrounded lengths, then round: 17.7 m.

Find the side in metres, then convert to the requested unit.

1. The known side AB=1.6 m is opposite \angle C=180^\circ-37^\circ-84^\circ=59^\circ.
2. AC is opposite \angle B=84^\circ. Sine rule: AC=\dfrac{1.6\sin84^\circ}{\sin59^\circ}=1.85638\ldots m.
3. There are 100 cm in 1 m, so AC=185.638\ldots cm.
4. AC=185.6 cm to 1 decimal place.

Find the angle opposite the shorter known side, then use the angle sum.

1. AC=14 is opposite the known angle \angle ABC=58^\circ, and AB=10 is opposite \angle ACB.
2. Sine rule: \dfrac{\sin C}{10}=\dfrac{\sin58^\circ}{14}, so \sin C=\dfrac{10\sin58^\circ}{14}=0.60574\ldots
3. \angle ACB=\sin^{-1}(0.60574\ldots)=37.282\ldots^\circ (acute, as stated).
4. Angles in a triangle add to 180^\circ: \angle BAC=180^\circ-58^\circ-37.282\ldots^\circ=84.717\ldots^\circ, which is 84.7^\circ to 1 decimal place.

After finding a sloping side, use a right-angled triangle to find the height.

1. \angle C=180^\circ-44^\circ-73^\circ=63^\circ, which is opposite the known side AB=15.
2. AC is opposite \angle B=73^\circ: AC=\dfrac{15\sin73^\circ}{\sin63^\circ}=16.099\ldots cm.
3. The perpendicular from C to AB makes a right-angled triangle with hypotenuse AC and angle 44^\circ at A, so the height is AC\sin44^\circ.
4. Height =16.099\ldots\times\sin44^\circ=11.183\ldots, which is 11.2 cm to 1 decimal place.

Use inverse sine for an angle before applying the sine rule again.

1. BC=9 is opposite the known angle \angle A=31^\circ, and AC=13 is opposite \angle B.
2. Sine rule: \sin B=\dfrac{13\sin31^\circ}{9}=0.74394\ldots, so the acute angle is \angle B=48.068\ldots^\circ.
3. Angles in a triangle: \angle C=180^\circ-31^\circ-48.068\ldots^\circ=100.931\ldots^\circ.
4. AB is opposite \angle C. Sine rule: AB=\dfrac{9\sin100.931\ldots^\circ}{\sin31^\circ}=17.157\ldots
5. AB=17.2 cm to 1 decimal place.

Convert the exterior angle to the interior angle before finding the remaining angle.

1. Angles on a straight line add to 180^\circ, so the interior angle \angle ABC=180^\circ-126^\circ=54^\circ.
2. Angles in a triangle: \angle ACB=180^\circ-49^\circ-54^\circ=77^\circ. This is opposite the known side AB=8.6.
3. BC is opposite \angle BAC=49^\circ. Sine rule: \dfrac{BC}{\sin49^\circ}=\dfrac{8.6}{\sin77^\circ}.
4. So BC=\dfrac{8.6\sin49^\circ}{\sin77^\circ}=6.661\ldots, which is 6.7 cm to 1 decimal place.

Use the exact values of \sin45^\circ and \sin60^\circ, then rationalise the denominator.

1. AC is opposite \angle ABC=60^\circ, and BC=6 is opposite \angle BAC=45^\circ.
2. Sine rule: AC=\dfrac{6\sin60^\circ}{\sin45^\circ}.
3. Exact values: \sin60^\circ=\dfrac{\sqrt3}{2} and \sin45^\circ=\dfrac{\sqrt2}{2}. The halves cancel, so AC=\dfrac{6\sqrt3}{\sqrt2}.
4. Rationalise the denominator: \dfrac{6\sqrt3}{\sqrt2}\times\dfrac{\sqrt2}{\sqrt2}=\dfrac{6\sqrt6}{2}=3\sqrt6.
5. So k=3.

Find the common side in the first triangle, then match it to its opposite angle in the second.

1. AC is a side of both triangles, so find it in triangle ABC first. \angle ACB=180^\circ-42^\circ-67^\circ=71^\circ, opposite the known side AB=10.
2. AC is opposite \angle ABC=67^\circ: AC=\dfrac{10\sin67^\circ}{\sin71^\circ}=9.7354\ldots cm.
3. Triangle ACD: \angle ADC=180^\circ-38^\circ-72^\circ=70^\circ, which is opposite AC.
4. CD is opposite \angle CAD=38^\circ: CD=\dfrac{9.7354\ldots\times\sin38^\circ}{\sin70^\circ}=6.378\ldots
5. CD=6.4 cm to 1 decimal place.

Find the fence length first, then find another side of the triangle whose area is required.

1. Triangle ABC: \angle ACB=180^\circ-56^\circ-74^\circ=50^\circ, opposite AB=18. The fence AC is opposite \angle ABC=74^\circ, so AC=\dfrac{18\sin74^\circ}{\sin50^\circ}=22.587\ldots m.
2. Triangle ACD: \angle ADC=180^\circ-47^\circ-61^\circ=72^\circ, which is opposite AC.
3. AD is opposite \angle ACD=61^\circ: AD=\dfrac{22.587\ldots\times\sin61^\circ}{\sin72^\circ}=20.771\ldots m.
4. AC and AD enclose \angle CAD=47^\circ, so area =\dfrac12\times AC\times AD\times\sin47^\circ.
5. Area =\dfrac12\times22.587\ldots\times20.771\ldots\times\sin47^\circ=171.565\ldots, which is 171.6 m^2 to 1 decimal place.