G22, G23 - Sine rule, cosine rule and triangle area (Higher) | AQA GCSE Mathematics

A triangle does not need a right angle for its sides and angles to be connected. G22 gives the sine and cosine rules for missing lengths and angles; G23 gives an area formula using two sides and the angle between them. You will choose a rule from the information available and recognise when an angle could have two possible values.

Using the sine rule for lengths

Two triangles each have sides $8\,\mathrm{cm}$ and $11\,\mathrm{cm}$ . One has an angle of $30^\circ$ between those sides; the other has $150^\circ$ . Could their areas be equal? They look very different, but a shared height will settle this by the end.

You already know

Angles in a triangle add to $180^\circ$ . In a right triangle, sine is opposite divided by hypotenuse. Use degree mode and keep unrounded values until the final answer.

For a general triangle, give each side the lower-case letter matching its opposite angle: side $a$ faces angle $A$ , side $b$ faces $B$ , and side $c$ faces $C$ . Opposite means across the triangle, not touching that vertex.

[DIAGRAM: asset_name: Match every side to its opposite angle; asset_slug: aqa8300_g21_opposites; caption: Lower-case side letters face their matching capital-letter angles; description: A non-right scalene triangle ABC. Side BC is labelled a and matches a blue angle arc A. Side AC is labelled b and matches a sage angle arc B. Side AB is labelled c and matches a clay angle arc C. A perpendicular height from C to AB is dashed and marked h, with a right-angle square at its foot.]
Lower-case side letters face their matching capital-letter angles

Drop a perpendicular height $h$ onto side $c$ . In an acute triangle, $h=b\sin A=a\sin B$ . Therefore $a/\sin A=b/\sin B$ . Repeating the argument with another height brings in $c/\sin C$ . The rule also holds for obtuse triangles, using a height outside the triangle.

Sine rule

In any triangle, with each side opposite its matching angle,

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

Use two fractions at a time. For a missing side, you need a known side with its opposite angle, plus the angle opposite the wanted side.

On the 2027 Higher formulae sheet.

The known opposite pair is your starting point. If the third angle is missing, work it out before substituting.

Worked example 1: two matching pairs

Question. Calculator allowed. In triangle $ABC$ , $A=42^\circ$ , $B=67^\circ$ and $a=8\,\mathrm{cm}$ . Work out $b$ to $3$ significant figures.

Plan. Match $8$ with $42^\circ$ , then use the sine rule for $b$ .

\begin{aligned} \frac{b}{\sin67^\circ}&=\frac8{\sin42^\circ} &&\text{pair opposite sides and angles}\\ b&=\frac{8\sin67^\circ}{\sin42^\circ} &&\text{multiply by }\sin67^\circ\\ &=11.00538334\ldots &&\text{degree-mode calculator} \end{aligned}

Why this step? A larger angle faces a longer side. Putting $\sin42^\circ$ on top would reverse that relationship.

Answer. $b=11.0\,\mathrm{cm}$ .

Check. $67^\circ>42^\circ$ , and $11.0053\ldots>8$ , as expected.

Change the information given: the known side is now opposite the angle you must find first.

Worked example 2: find the third angle first

Question. Calculator allowed. In triangle $ABC$ , $A=46^\circ$ , $B=71^\circ$ and $c=12\,\mathrm{cm}$ . Work out $a$ to $3$ significant figures.

Plan. Find $C$ using the angle sum, then match side $c$ with it.

\begin{aligned} C&=180^\circ-46^\circ-71^\circ=63^\circ &&\text{triangle angle sum}\\ \frac{a}{\sin46^\circ}&=\frac{12}{\sin63^\circ} &&\text{sine rule}\\ a&=\frac{12\sin46^\circ}{\sin63^\circ}=9.688007180\ldots &&\text{rearrange} \end{aligned}

Why this step? The $12$ belongs with $C$ , not whichever angle happens to be written next to it in the question.

Answer. $a=9.69\,\mathrm{cm}$ .

Check. $46^\circ<63^\circ$ , so $a<12$ , as found.

Mark matching letters on a sketch if the fraction arrangement feels slippery.

Watch out

Wrong: pair side $a$ with angle $B$ because they touch. Right: pair a side with the angle opposite it. The sine rule compares opposite pairs.

Try a different known opposite pair.

Using the sine rule for angles

When the angle is unknown, putting the sine terms on top often makes rearranging easier. The reciprocal form $\sin A/a=\sin B/b$ is the same sine rule, not a second result to memorise.

An important boundary appears here. Sine gives the same value for supplementary angles: for example, $\sin30^\circ=\sin150^\circ=1/2$ . A height above a base can be the same when a side leans forwards or backwards. So an inverse-sine answer may not be the only possible triangle angle. This supplementary-angle fact is not on the formulae sheet: learn it.

Worked example 3: an acute angle is specified

Question. Calculator allowed. Triangle $ABC$ has $a=8\,\mathrm{cm}$ , $b=11\,\mathrm{cm}$ , $A=35^\circ$ , and acute angle $B$ . Work out $B$ to $1$ decimal place.

Plan. Rearrange the sine rule for $\sin B$ , then take inverse sine.

\begin{aligned} \frac{\sin B}{11}&=\frac{\sin35^\circ}{8} &&\text{reciprocal sine rule}\\ \sin B&=\frac{11\sin35^\circ}{8}=0.788667599\ldots &&\text{multiply by }11\\ B&=\sin^{-1}(11\sin35^\circ/8)=52.06117040\ldots^\circ &&\text{acute solution} \end{aligned}

Why this step? The question states that $B$ is acute, so the calculator's acute answer is the required one.

Answer. $B=52.1^\circ$ .

Check. $B>A$ , consistent with $b>a$ . The remaining angle is positive.

Change one condition: the wanted angle is obtuse. The calculator still returns its acute partner first.

Worked example 4: choosing the obtuse angle

Question. Calculator allowed. Triangle $ABC$ has $a=7\,\mathrm{cm}$ , $b=10\,\mathrm{cm}$ , $A=30^\circ$ , and obtuse angle $B$ . Work out $B$ to $1$ decimal place.

Plan. Find the acute inverse-sine angle, then subtract it from $180^\circ$ .

\begin{aligned} \sin B&=\frac{10\sin30^\circ}{7}=\frac57 &&\text{sine rule}\\ \beta&=\sin^{-1}(5/7)=45.58469140\ldots^\circ &&\text{acute partner}\\ B&=180^\circ-\beta=134.4153085\ldots^\circ &&\text{choose the obtuse solution} \end{aligned}

Why this step? Sine cannot distinguish these two supplementary angles. “Obtuse” chooses the one above $90^\circ$ .

Answer. $B=134.4^\circ$ .

Check. $30^\circ+134.4153\ldots^\circ<180^\circ$ , so a positive third angle remains. Both partners have sine $5/7$ .

If no acute/obtuse condition is given, check both candidates against the triangle's other angles and stated geometry. There may be two possible triangles. If $\sin B>1$ , there is no possible triangle; if the candidate angles make the angle sum at least $180^\circ$ , reject them.

Watch out

Wrong: “Inverse sine always gives the triangle's angle.” Right: inverse sine gives the acute partner for a positive ratio. Check whether an obtuse supplementary angle also fits the question.

The next question removes the ambiguity by specifying an acute angle.

Using the cosine rule for lengths

Suppose you know two sides and the angle between them. There is no known opposite side-angle pair, so the sine rule cannot start. The included angle is the angle between the two known sides.

Cosine rule

In any triangle,

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

Angle $A$ is between sides $b$ and $c$ , and side $a$ is opposite $A$ . Relabel the triangle to suit the wanted side.

On the 2027 Higher formulae sheet.

When $A=90^\circ$ , $\cos A=0$ , so this becomes Pythagoras. When $A$ is acute, the correction is subtracted; when it is obtuse, cosine is negative and the opposite side becomes longer. The cosine rule extends the right-triangle relationship rather than replacing it.

Worked example 5: an acute included angle

Question. Calculator allowed. Two sides of a triangle are $7\,\mathrm{cm}$ and $9\,\mathrm{cm}$ . The angle between them is $60^\circ$ . Work out the third side $x$ to $3$ significant figures.

Plan. Use the cosine rule with $x$ opposite the given angle, then take the square root.

\begin{aligned} x^2&=7^2+9^2-2(7)(9)\cos60^\circ &&\text{cosine rule}\\ &=130-63=67 &&\text{use }\cos60^\circ=1/2\\ x&=\sqrt{67}=8.185352771\ldots &&\text{positive square root} \end{aligned}

Why this step? The result before the square root is $x^2$ . The angle must be the one between $7$ and $9$ .

Answer. $x=8.19\,\mathrm{cm}$ .

Check. $9-7<x<9+7$ , so the triangle inequality is satisfied.

Change only the included angle to $120^\circ$ . A wider opening should produce a longer opposite side.

Worked example 6: an obtuse included angle

Question. Calculator allowed. Two sides are $7\,\mathrm{cm}$ and $9\,\mathrm{cm}$ , with included angle $120^\circ$ . Work out the opposite side $x$ to $3$ significant figures.

Plan. Keep the full signed cosine term inside the square root.

\begin{aligned} x^2&=7^2+9^2-2(7)(9)\cos120^\circ &&\text{cosine rule}\\ &=130-126(-1/2)=193 &&\text{cosine is negative}\\ x&=\sqrt{193}=13.89244398\ldots &&\text{positive square root} \end{aligned}

Why this step? Subtracting a negative adds. The obtuse triangle's longest side must be opposite $120^\circ$ .

Answer. $x=13.9\,\mathrm{cm}$ .

Check. It is longer than $9$ but shorter than $7+9=16$ , as required.

Use brackets around a negative cosine value if entering it separately.

Watch out

Wrong: $x^2=b^2+c^2-2bc\sin A$ . Right: the correction in the cosine rule uses cosine. Sine belongs in the sine rule and the area formula, not in this correction.

Apply the same rule to a new obtuse angle.

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G22, G23 - Sine rule, cosine rule and triangle area (Higher)

Using the sine rule for lengths

Using the sine rule for angles

Using the cosine rule for lengths

Continue this lesson.

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