Sine rule, cosine rule and triangle area

The sine rule, cosine rule and triangle area formula solve triangles that do not contain a right angle. Success depends less on memorising formulae than on labelling opposite pairs, recognising the information given, and deciding whether one or two triangles are possible.

Throughout this lesson, side $a$ is opposite angle $A$, side $b$ is opposite angle $B$, and side $c$ is opposite angle $C$.

$$\boxed{a\leftrightarrow A,\qquad b\leftrightarrow B,\qquad c\leftrightarrow C}$$

Capital letters label angles and the matching lower-case letters label opposite sides. This convention is essential: $a$ need not be adjacent to $A$.

Prerequisites

You should be able to:

• use Pythagoras and trigonometry in right-angled triangles;
• rearrange formulae and evaluate inverse trigonometric functions;
• use exact values from exact trigonometric values;
• understand that angles in a triangle sum to $180^\circ$;
• keep full calculator values until the final answer.

If these rules are new, the gentler foundations lesson on sine and cosine rules is a useful starting point.

Choose the rule from the information

Mark every known side and angle before calculating. Then identify the pattern.

Information available	Required quantity	Best starting point
An opposite side and angle pair, plus another side or angle	Side or angle	Sine rule
Two sides and their included angle, SAS	Third side	Cosine rule
Three sides, SSS	Angle	Cosine rule
Two sides and their included angle	Area	$\frac12 ab\sin C$
Two sides and a non-included angle, SSA	Triangle	Sine rule, checking the ambiguous case

The included angle is the angle physically between the two stated sides. For sides $a$ and $b$, it is $C$.

The quick decision

Look first for a complete opposite pair. If one exists, the sine rule is usually available. If no pair exists but you know SAS or SSS, use the cosine rule.

The sine rule

For any triangle,

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

Use the reciprocal form when finding an angle:

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

Choose only the two fractions needed. Every numerator must remain paired with its opposite angle.

Why the sine rule works

Drop a perpendicular of height $h$ from $C$ to side $c$. In the two right-angled triangles,

$$\sin A=\frac{h}{b} \quad\Longrightarrow\quad h=b\sin A,$$

and

$$\sin B=\frac{h}{a} \quad\Longrightarrow\quad h=a\sin B.$$

Equating the two expressions for the same height gives

$$b\sin A=a\sin B.$$

Dividing by $\sin A\sin B$ gives

$$\frac{a}{\sin A}=\frac{b}{\sin B}.$$

Repeating the argument with another perpendicular includes the third pair.

Worked example 1: find a side

In triangle $ABC$,

$$A=48^\circ,\qquad B=71^\circ,\qquad a=9.4\text{ cm}.$$

Find $b$.

The known opposite pair is $a=9.4$ and $A=48^\circ$. Pair $b$ with $B=71^\circ$:

$$\frac{b}{\sin71^\circ}=\frac{9.4}{\sin48^\circ}.$$

Therefore

$$\begin{aligned} b &=\frac{9.4\sin71^\circ}{\sin48^\circ}\\ &=11.959\ldots \end{aligned}$$

so

$$\boxed{b=12.0\text{ cm}}\qquad\text{to 3 significant figures}.$$

The larger side $b$ lies opposite the larger angle $B$, which supports the answer.

Worked example 2: find an angle

In triangle $PQR$, side $p=7.2$ cm is opposite $P=35^\circ$, and side $q=9.1$ cm is opposite $Q$. Find the possible values of $Q$.

Use the angle-over-side form:

$$\frac{\sin Q}{9.1}=\frac{\sin35^\circ}{7.2}.$$

Hence

$$\sin Q=\frac{9.1\sin35^\circ}{7.2}=0.7249\ldots$$

The calculator gives the acute solution

$$Q_1=\sin^{-1}(0.7249\ldots)=46.46\ldots^\circ.$$

But sine is also positive in the second quadrant, so

$$Q_2=180^\circ-Q_1=133.53\ldots^\circ.$$

Both satisfy the triangle angle sum because

$$35^\circ+46.46\ldots^\circ<180^\circ$$

and

$$35^\circ+133.53\ldots^\circ<180^\circ.$$

Therefore

$$\boxed{Q=46.5^\circ\text{ or }133.5^\circ}$$

to $1$ decimal place. This is the ambiguous sine case, examined fully below.

Self check 1

In triangle $ABC$, $A=42^\circ$, $C=83^\circ$ and $c=14$ cm.

1. Find $B$.
2. Find $a$.

Answer

$$B=180^\circ-42^\circ-83^\circ=\boxed{55^\circ}.$$

Using the opposite pairs $(a,A)$ and $(c,C)$,

$$\frac{a}{\sin42^\circ}=\frac{14}{\sin83^\circ}.$$

Thus

$$a=\frac{14\sin42^\circ}{\sin83^\circ}=9.437\ldots$$

and $\boxed{a=9.44\text{ cm}}$ to $3$ significant figures.

The ambiguous sine case

An inverse sine calculation returns only its principal value between $-90^\circ$ and $90^\circ$. Yet for $0^\circ<\theta<180^\circ$,

$$\sin\theta=\sin(180^\circ-\theta).$$

Consequently, SSA information can describe two triangles, one triangle, or no triangle. The issue arises when an angle and its opposite side are known, together with a second side, but the angle between the two known sides is not known.

Suppose $A$, $a$ and $b$ are given, where $A$ is acute. Define the perpendicular height

$$h=b\sin A.$$

Then:

Comparison	Number of triangles
$a<h$	none
$a=h$	one right-angled triangle
$h<a<b$	two
$a\geq b$	one

If the given angle $A$ is obtuse, its opposite side must be the longest. Therefore there is one triangle if $a>b$, and no triangle if $a\leq b$.

A reliable method

When the sine rule gives an angle $B_1$:

1. calculate the principal value $B_1$;
2. calculate $B_2=180^\circ-B_1$;
3. test both using $A+B<180^\circ$;
4. complete every valid triangle separately;
5. reject impossible angles explicitly.

Do not add the second angle automatically

The supplementary value has the same sine, but it is valid only if it leaves a positive third angle. Context may also remove a solution.

Worked example 3: two complete triangles

In triangle $ABC$,

$$A=30^\circ,\qquad a=7\text{ cm},\qquad b=10\text{ cm}.$$

Find all possible values of $B$ and $c$.

First,

$$\frac{\sin B}{10}=\frac{\sin30^\circ}{7},$$

so

$$\sin B=\frac{10\sin30^\circ}{7}=\frac57.$$

The two candidates are

$$B_1=\sin^{-1}\left(\frac57\right)=45.584\ldots^\circ$$

and

$$B_2=180^\circ-B_1=134.415\ldots^\circ.$$

Both are possible because $30^\circ+B_i<180^\circ$.

For the first triangle,

$$C_1=180^\circ-30^\circ-45.584\ldots^\circ=104.415\ldots^\circ.$$

Then

$$c_1=\frac{7\sin C_1}{\sin30^\circ}=13.559\ldots\text{ cm}.$$

For the second triangle,

$$C_2=180^\circ-30^\circ-134.415\ldots^\circ=15.584\ldots^\circ,$$

so

$$c_2=\frac{7\sin C_2}{\sin30^\circ}=3.761\ldots\text{ cm}.$$

Thus the two solutions are

$$\boxed{B=45.6^\circ,\quad c=13.6\text{ cm}}$$

or

$$\boxed{B=134.4^\circ,\quad c=3.76\text{ cm}}.$$

The cosine rule

For any triangle,

$$\boxed{a^2=b^2+c^2-2bc\cos A}.$$

The cyclic forms follow by relabelling:

$$b^2=c^2+a^2-2ca\cos B,$$ $$c^2=a^2+b^2-2ab\cos C.$$

The isolated angle form is often safer for SSS questions:

$$\boxed{\cos A=\frac{b^2+c^2-a^2}{2bc}}.$$

Notice the structure: the side alone on the left is opposite the angle in the cosine. The other two sides appear in the product.

Why the cosine rule works

Place $B=(0,0)$, $C=(a,0)$ and $A=(c\cos B,c\sin B)$. The squared distance from $A$ to $C$ is $b^2$, so

$$\begin{aligned} b^2 &=(a-c\cos B)^2+(c\sin B)^2\\ &=a^2-2ac\cos B+c^2(\cos^2B+\sin^2B)\\ &=a^2+c^2-2ac\cos B. \end{aligned}$$

This is the cosine rule. If $B=90^\circ$, then $\cos B=0$ and it reduces to

$$b^2=a^2+c^2,$$

so Pythagoras’ theorem is its right-angled special case.

Worked example 4: SAS, find a side

Two sides of a triangle are $8.2$ cm and $11.5$ cm, and their included angle is $63^\circ$. Find the opposite side $x$.

Because the included angle is known, use the cosine rule:

$$\begin{aligned} x^2 &=8.2^2+11.5^2-2(8.2)(11.5)\cos63^\circ\\ &=113.873\ldots \end{aligned}$$

Lengths are positive, so

$$x=\sqrt{113.873\ldots}=10.671\ldots$$

and

$$\boxed{x=10.7\text{ cm}}$$

to $3$ significant figures.

Worked example 5: SSS, find the largest angle

A triangle has side lengths $6$ cm, $8$ cm and $11$ cm. Find its largest angle.

The largest angle lies opposite the longest side, $11$ cm. Call it $C$. Then

$$\begin{aligned} \cos C &=\frac{6^2+8^2-11^2}{2(6)(8)}\\ &=-\frac{21}{96}. \end{aligned}$$

Therefore

$$C=\cos^{-1}\left(-\frac{21}{96}\right)=102.635\ldots^\circ,$$

so

$$\boxed{C=102.6^\circ}.$$

The negative cosine correctly indicates an obtuse angle.

Classifying a triangle without finding its angles

Let $a$ be the longest side. Comparing $a^2$ with $b^2+c^2$ classifies the opposite angle $A$:

$$\begin{array}{c|c} a^2<b^2+c^2 & A<90^\circ\\ a^2=b^2+c^2 & A=90^\circ\\ a^2>b^2+c^2 & A>90^\circ \end{array}$$

This follows from the sign of $\cos A$ in the cosine rule.

Self check 2

A triangle has side lengths $5$ cm, $7$ cm and $9$ cm.

1. Is its largest angle acute, right or obtuse?
2. Find that angle.

Answer

The longest side is $9$ cm. Since

$$9^2=81>5^2+7^2=74,$$

the largest angle is obtuse.

If that angle is $C$, then

$$\cos C=\frac{5^2+7^2-9^2}{2(5)(7)}=-\frac1{10}.$$

Therefore

$$\boxed{C=\cos^{-1}(-0.1)=95.7^\circ}$$

to $1$ decimal place.

Area of a triangle

If two sides and their included angle are known, then

$$\boxed{K=\frac12 ab\sin C}$$

where $K$ denotes area. Equivalent forms are

$$K=\frac12 bc\sin A=\frac12 ca\sin B.$$

To derive the formula, take side $a$ as the base. The perpendicular height is $b\sin C$, so

$$K=\frac12\times a\times b\sin C.$$

The angle must be included between the two sides used.

Worked example 6: area from two sides and an angle

Find the area of a triangle with sides $12.4$ m and $17.1$ m enclosing an angle of $38^\circ$.

$$\begin{aligned} K &=\frac12(12.4)(17.1)\sin38^\circ\\ &=65.272\ldots \end{aligned}$$

Therefore

$$\boxed{K=65.3\text{ m}^2}$$

to $3$ significant figures. Area requires square units.

Worked example 7: find an angle from an area

A triangle has area $24\text{ cm}^2$. Two of its sides have lengths $8$ cm and $9$ cm. Find the possible included angles $C$.

$$24=\frac12(8)(9)\sin C,$$

so

$$\sin C=\frac{24}{36}=\frac23.$$

Thus

$$C_1=\sin^{-1}\left(\frac23\right)=41.810\ldots^\circ$$

or

$$C_2=180^\circ-C_1=138.189\ldots^\circ.$$

Both are valid included angles, so

$$\boxed{C=41.8^\circ\text{ or }138.2^\circ}.$$

Equal supplementary sines explain why the same two sides can enclose two different angles but produce the same area.

Exact triangle calculations

Do not replace exact data by decimals unless the question requests an approximation.

Worked example 8: exact area and side

Sides $a=4$ and $b=7$ enclose angle $C=60^\circ$. Find the area and side $c$ exactly.

For the area,

$$K=\frac12(4)(7)\sin60^\circ =14\cdot\frac{\sqrt3}{2} =\boxed{7\sqrt3}.$$

For the third side,

$$\begin{aligned} c^2 &=4^2+7^2-2(4)(7)\cos60^\circ\\ &=16+49-56\left(\frac12\right)\\ &=37. \end{aligned}$$

Hence

$$\boxed{c=\sqrt{37}}.$$

Multi-step triangle problems

One rule may create the information needed by another. Keep a diagram labelled, show unrounded intermediate values, and ask what the next rule requires.

Worked example 9: solve a triangle fully

In triangle $ABC$,

$$b=12\text{ cm},\qquad c=16\text{ cm},\qquad A=52^\circ.$$

Find $a$, $B$ and the area.

There is no complete opposite pair initially. The two known sides enclose $A$, so begin with the cosine rule:

$$\begin{aligned} a^2 &=12^2+16^2-2(12)(16)\cos52^\circ\\ &=163.586\ldots, \end{aligned}$$

giving

$$a=12.790\ldots\text{ cm}.$$

Now use the sine rule. Since $b=12<a=12.790\ldots$, its opposite angle $B$ must be smaller than $A=52^\circ$:

$$\frac{\sin B}{12}=\frac{\sin52^\circ}{12.790\ldots}.$$

Therefore

$$B=47.67\ldots^\circ.$$

The supplementary value is impossible because it would exceed $52^\circ$ despite lying opposite the shorter side. Finally,

$$\begin{aligned} K &=\frac12(12)(16)\sin52^\circ\\ &=75.648\ldots\text{ cm}^2. \end{aligned}$$

Thus

$$\boxed{a=12.8\text{ cm},\quad B=47.7^\circ,\quad K=75.6\text{ cm}^2}.$$

Bearings and modelling

A bearing is measured clockwise from north and written with three digits, such as $065^\circ$. The angle inside a triangle is often not the bearing itself. Use parallel north lines, angles around a point, or a sketch to derive the interior angle first.

Worked example 10: distances and bearings

A boat sails $18$ km from $P$ on a bearing of $040^\circ$ to $Q$. It then sails $25$ km from $Q$ on a bearing of $130^\circ$ to $R$. Find the direct distance $PR$.

The bearing from $Q$ back to $P$ is

$$040^\circ+180^\circ=220^\circ.$$

At $Q$, the interior angle between directions $220^\circ$ and $130^\circ$ is

$$220^\circ-130^\circ=90^\circ.$$

The cosine rule gives

$$\begin{aligned} PR^2 &=18^2+25^2-2(18)(25)\cos90^\circ\\ &=949. \end{aligned}$$

Hence

$$\boxed{PR=\sqrt{949}=30.8\text{ km}}$$

to $3$ significant figures.

Model first, calculate second

In a worded problem, distinguish the physical information from the triangle information. Draw the route, add north lines where needed, calculate the interior angle, then select a rule.

Common misconceptions

Pairing an angle with an adjacent side

In the sine rule, each angle pairs with the side opposite it. Write the matching letters before substituting:

$$\frac{a}{\sin A}=\frac{b}{\sin B}.$$

Using the sine rule without a complete pair

SAS data contain no known opposite side and angle pair. Find the third side with the cosine rule first.

Using a non-included angle in the cosine or area formula

In

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

and

$$K=\frac12bc\sin A,$$

$A$ is the angle between $b$ and $c$.

Missing the second sine solution

If $\sin B=k$, test both

$$B=\sin^{-1}k \quad\text{and}\quad B=180^\circ-\sin^{-1}k.$$

Never assume either is valid until the angle sum is checked.

Rounding too early

Store calculator values or carry at least several extra figures. Premature rounding can alter a final answer, especially in a long chain of triangle calculations.

Ignoring whether an answer is geometrically possible

Always check:

• the largest angle is opposite the largest side;
• all angles are positive and total $180^\circ$;
• the sum of any two sides exceeds the third side;
• $|b-c|<a<b+c$ for a third side $a$;
• lengths and areas have appropriate units.

Mixed self check

1. Two sides of a triangle are $7$ cm and $13$ cm, enclosing an angle of $104^\circ$. Find the third side.
2. A triangle has sides $8$ cm, $10$ cm and $14$ cm. Find its largest angle.
3. Find the area of a triangle with sides $9$ cm and $15$ cm enclosing $32^\circ$.
4. In triangle $ABC$, $A=38^\circ$, $a=8$ cm and $b=11$ cm. Find all possible values of $B$.
5. Explain why no triangle exists when $A=30^\circ$, $a=4$ cm and $b=10$ cm.

Answers

1. By the cosine rule,

   $$x^2=7^2+13^2-2(7)(13)\cos104^\circ,$$

   so $\boxed{x=16.2\text{ cm}}$ to $3$ significant figures.

2. The largest angle $C$ is opposite $14$ cm:

   $$\cos C=\frac{8^2+10^2-14^2}{2(8)(10)}=-0.2.$$

   Hence $\boxed{C=101.5^\circ}$ to $1$ decimal place.

$$K=\frac12(9)(15)\sin32^\circ=\boxed{35.8\text{ cm}^2}$$

to $3$ significant figures.

$$\sin B=\frac{11\sin38^\circ}{8},$$

giving $B_1=57.8^\circ$ and $B_2=122.2^\circ$. Both leave a positive third angle, so $\boxed{B=57.8^\circ\text{ or }122.2^\circ}$.

5. The perpendicular height is

   $$h=b\sin A=10\sin30^\circ=5\text{ cm}.$$

   Since $a=4<h=5$, side $a$ cannot reach the base to close the triangle. Therefore no triangle exists.

Summary

For opposite pairs $a\leftrightarrow A$, $b\leftrightarrow B$ and $c\leftrightarrow C$:

$$\boxed{\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}}$$ $$\boxed{a^2=b^2+c^2-2bc\cos A}$$ $$\boxed{K=\frac12bc\sin A}.$$

Use the sine rule when a known opposite pair is available. Use the cosine rule for SAS or SSS. Use the area formula with two sides and their included angle. Whenever inverse sine is used with SSA data, test the supplementary angle.

Next steps

• Develop algebraic formulae for compound and double angles.
• Apply triangle reasoning in trigonometric proof and modelling.
• Connect triangle ratios to directions and magnitudes in vectors.
• Use closed force triangles in resolving forces.