Sine rule, cosine rule and triangle area

The sine rule, cosine rule and triangle area formula extend trigonometry to triangles without a right angle. The difficult part is usually not substitution. It is matching each angle with its opposite side and choosing a formula from the information given.

Prerequisites

You should be able to:

• label sides and angles in a triangle;
• use sine, cosine and inverse trigonometric functions;
• rearrange equations and work with squares and square roots;
• use the angle sum $A+B+C=180^\circ$;
• keep full calculator accuracy until the final answer.

Review Pythagoras and right-angled trigonometry, rearranging formulae or calculator fluency if necessary.

Standard triangle notation

In triangle $ABC$, lower-case letters name the opposite sides:

$$a=BC,\qquad b=CA,\qquad c=AB.$$

Thus $a$ is opposite angle $A$, $b$ is opposite $B$, and $c$ is opposite $C$. An angle and its opposite side form an opposite pair.

Label opposite pairs first

Before selecting a rule, mark each known angle and trace across the triangle to its opposite side. Never match an angle with a side merely because they share a vertex.

Choosing the correct formula

Use the information in the question, not the appearance of the diagram.

Information available	Typical goal	Method
An opposite angle and side pair, plus another side or angle	Find the fourth value	Sine rule
Two sides and their included angle	Find the third side	Cosine rule
Three sides	Find an angle	Cosine rule
Two sides and their included angle	Find area	$\frac12 ab\sin C$

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

The sine rule

For any triangle $ABC$,

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

The reciprocal form is often convenient when finding an angle:

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

Use only the two fractions needed. Every fraction must contain a correctly matched opposite pair.

Worked example: find a side

In triangle $ABC$, $A=48^\circ$, $B=71^\circ$ and $a=9.4$ cm. Find $b$.

The known pair is $A=48^\circ$ and $a=9.4$:

$$\begin{aligned} \frac{b}{\sin71^\circ}&=\frac{9.4}{\sin48^\circ}\\ b&=\frac{9.4\sin71^\circ}{\sin48^\circ}\\ &=11.96\ldots\\ &=12.0\text{ cm}\quad\text{to 3 s.f.} \end{aligned}$$

Since $71^\circ>48^\circ$, its opposite side should be longer: $b>9.4$. This confirms the answer is plausible.

Worked example: find an angle

In a triangle, $a=7.2$ cm, $b=10.5$ cm and $A=34^\circ$. Find the acute possible value of $B$.

$$\begin{aligned} \frac{\sin B}{10.5}&=\frac{\sin34^\circ}{7.2}\\ \sin B&=\frac{10.5\sin34^\circ}{7.2}\\ B&=\sin^{-1}\left(\frac{10.5\sin34^\circ}{7.2}\right)\\ &=54.6^\circ\quad\text{to 1 d.p.} \end{aligned}$$

This calculation may have a second valid answer. That issue is addressed below.

Why the sine rule works

Drop a perpendicular of height $h$ from $C$ to $AB$. Using the two right-angled triangles,

$$h=b\sin A \qquad\text{and}\qquad h=a\sin B.$$

Therefore $b\sin A=a\sin B$, so

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

Repeating the argument with another perpendicular gives all three equal ratios. The common value is also the diameter of the triangle’s circumcircle, a fact that becomes useful in more advanced geometry.

The ambiguous sine case

For angles between $0^\circ$ and $180^\circ$,

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

Consequently, inverse sine returns one angle but there may be a second:

$$\theta_2=180^\circ-\theta_1.$$

This ambiguity can occur when two sides and a non-included angle are known. It does not occur when finding a side, or when the angle is fixed by three known sides.

Worked example: two possible triangles

In triangle $ABC$, $A=35^\circ$, $a=8$ cm and $b=11$ cm. Find all possible values of $B$ and $C$.

First use the sine rule:

$$\sin B=\frac{11\sin35^\circ}{8}=0.78867\ldots$$

The calculator gives

$B_1=52.1^\circ.$

The supplementary angle is

$B_2=180^\circ-52.1^\circ=127.9^\circ.$

Both leave a positive third angle:

$$\begin{aligned} C_1&=180^\circ-35^\circ-52.1^\circ=92.9^\circ,\\ C_2&=180^\circ-35^\circ-127.9^\circ=17.1^\circ. \end{aligned}$$

So both triangles are possible, subject to rounding.

When the second angle is impossible

Suppose inverse sine gives $B_1=68^\circ$ and another known angle is $A=47^\circ$. The alternative is $B_2=112^\circ$, but

$47^\circ+112^\circ=159^\circ<180^\circ,$

so it is still possible. If instead $A=75^\circ$, then $75^\circ+112^\circ>180^\circ$, so the alternative must be rejected.

Always test the supplement

Do not automatically give two answers. Form the supplementary angle, then check that the three angles can be positive and sum to $180^\circ$.

The cosine rule

For any triangle $ABC$,

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

The other versions follow by cycling the letters:

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

The side alone on the left is opposite the angle in the cosine term. The other two sides multiply in the final term.

If $A=90^\circ$, then $\cos A=0$, so the formula becomes $a^2=b^2+c^2$. Pythagoras’ theorem is therefore a special case of the cosine rule.

Worked example: find a side

Two sides of a triangle are $7$ cm and $12$ cm, and their included angle is $58^\circ$. Find the opposite side $x$.

$$\begin{aligned} x^2&=7^2+12^2-2(7)(12)\cos58^\circ\\ &=103.973\ldots\\ x&=\sqrt{103.973\ldots}\\ &=10.2\text{ cm}\quad\text{to 3 s.f.} \end{aligned}$$

Use the positive square root because $x$ is a length. Also, $|12-7|<x<12+7$, so $5<x<19$ as required by the triangle inequality.

Worked example: find an angle from three sides

A triangle has sides $a=13$ cm, $b=9$ cm and $c=7$ cm. Find angle $A$.

Start with the version containing $A$ and rearrange before substituting:

$$\begin{aligned} a^2&=b^2+c^2-2bc\cos A\\ 2bc\cos A&=b^2+c^2-a^2\\ \cos A&=\frac{b^2+c^2-a^2}{2bc}. \end{aligned}$$

Therefore

$$\begin{aligned} A&=\cos^{-1}\left(\frac{9^2+7^2-13^2}{2(9)(7)}\right)\\ &=108.0^\circ\quad\text{to 1 d.p.} \end{aligned}$$

The longest side is $13$, so its opposite angle must be the largest. An obtuse result is reasonable.

Classifying a triangle from its sides

For the angle $A$ opposite the longest side $a$:

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

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

Area of a triangle

If two sides and their included angle are known,

$$\boxed{\text{Area}=\frac12 ab\sin C}.$$

Equivalent forms are $\frac12 bc\sin A$ and $\frac12 ca\sin B$. The angle must be between the two sides used.

Why the formula works

Taking $a$ as the base, the perpendicular height is $b\sin C$. Hence

$$\text{Area}=\frac12\times a\times b\sin C.$$

The formula remains valid for an obtuse included angle because $\sin C=\sin(180^\circ-C)$, which gives the same perpendicular height.

Worked example: direct area

Find the area of a triangle with sides $8.5$ cm and $13.2$ cm enclosing an angle of $42^\circ$.

$$\begin{aligned} \text{Area} &=\frac12(8.5)(13.2)\sin42^\circ\\ &=37.5\text{ cm}^2\quad\text{to 3 s.f.} \end{aligned}$$

Area uses square units.

Worked example: area from three sides

A triangle has sides $6$ cm, $8$ cm and $11$ cm. Find its area.

First find the angle $C$ between the sides $6$ and $8$, opposite side $11$:

$$\cos C=\frac{6^2+8^2-11^2}{2(6)(8)}=-\frac7{32}.$$

Then

$$\begin{aligned} \text{Area} &=\frac12(6)(8)\sin\left(\cos^{-1}\left(-\frac7{32}\right)\right)\\ &=23.4\text{ cm}^2\quad\text{to 3 s.f.} \end{aligned}$$

Keeping the exact fraction inside the calculator avoids compounding rounding error.

Solving a complete triangle

To solve a triangle means to find all unknown sides and angles.

Worked example: combine the rules

In triangle $ABC$, $b=12$ cm, $c=17$ cm and $A=63^\circ$. Find $a$, $B$ and the area.

The given angle is included between the two known sides, so begin with the cosine rule:

$$\begin{aligned} a^2&=12^2+17^2-2(12)(17)\cos63^\circ,\\ a&=15.740\ldots\text{ cm}. \end{aligned}$$

Now use the sine rule to find $B$:

$$\begin{aligned} \frac{\sin B}{12}&=\frac{\sin63^\circ}{15.740\ldots},\\ B&=42.8^\circ\quad\text{to 1 d.p.} \end{aligned}$$

There is no valid supplementary answer because $180^\circ-42.8^\circ+63^\circ>180^\circ$. Finally,

$$\text{Area}=\frac12(12)(17)\sin63^\circ=90.9\text{ cm}^2\quad\text{to 3 s.f.}$$

The remaining angle is $C=180^\circ-63^\circ-42.8^\circ=74.2^\circ$, subject to rounding.

Common misconceptions

Mismatching opposite pairs

In $a/\sin A=b/\sin B$, each side sits with its opposite angle. Writing $a/\sin B$ destroys the geometric relationship.

Using the sine rule without a known pair

Two sides and their included angle do not provide a complete opposite pair. Use the cosine rule to find the third side first.

Using a non-included angle in the area formula

In $\frac12 ab\sin C$, angle $C$ must lie between sides $a$ and $b$.

Forgetting the ambiguous case

When inverse sine finds an angle, test $180^\circ-\theta$. Inverse cosine for a triangle angle has only one value between $0^\circ$ and $180^\circ$.

Rounding intermediate values

Use the calculator’s stored value or retain the full expression. Round once, on the final line, to the requested accuracy.

Self-check

1. In triangle $ABC$, $A=46^\circ$, $a=8$ cm and $B=79^\circ$. Find $b$ to three significant figures.
2. Two sides have lengths $9$ cm and $14$ cm with included angle $105^\circ$. Find the third side to three significant figures.
3. A triangle has sides $5$ cm, $7$ cm and $9$ cm. Find the angle opposite the $9$ cm side to one decimal place.
4. Find the area of a triangle with sides $10$ cm and $16$ cm enclosing an angle of $38^\circ$.
5. In triangle $ABC$, $A=30^\circ$, $a=7$ cm and $b=10$ cm. Find all possible values of $B$ to one decimal place.
6. Explain which rule you would use first when given all three sides.

Answers

1. $b=8\sin79^\circ/\sin46^\circ=10.9$ cm.
2. $x=\sqrt{9^2+14^2-2(9)(14)\cos105^\circ}=18.5$ cm.
3. $\theta=\cos^{-1}((5^2+7^2-9^2)/(2(5)(7)))=95.7^\circ$.
4. $\frac12(10)(16)\sin38^\circ=49.3\text{ cm}^2$.
5. $\sin B=10\sin30^\circ/7=5/7$. Therefore $B=45.6^\circ$ or $134.4^\circ$. Both are possible because each leaves a positive third angle.
6. Use the cosine rule to find an angle. The sine rule cannot begin because no angle and opposite side pair is known.

What to learn next

You should now be able to select and apply all three non-right-angled triangle formulas, detect a second possible triangle, and combine methods without premature rounding.

Next, study:

• triangles and trigonometric rules for A-level extensions and modelling;
• radians for measuring angles independently of degrees;
• circle geometry for chords, cyclic figures and circumcircles;
• vectors for geometric arguments using magnitude and direction.