Sine Rule and Cosine Rule

A triangle does not need a right angle for its sides and angles to be related. The sine rule connects opposite side-angle pairs. The cosine rule connects three sides with an included angle. Together they let us solve triangles that right-triangle ratios cannot address directly.

The first decision is which information is known. Two angles and a side, two sides and their included angle, and three sides lead to different routes. Two sides and a non-included angle need special care because they can describe two different triangles.

On this page

Label opposite side-angle pairs

Use capital letters $A,B,C$ for angle measures and lowercase $a,b,c$ for the lengths opposite them. Side $a$ is opposite $A$ , not necessarily beside it. A consistent diagram prevents a correct formula from being applied to the wrong pair.

For an ordinary non-degenerate plane triangle, all sides are positive, all interior angles lie between $0^\circ$ and $180^\circ$ , and $A+B+C=180^\circ$ . The longest side lies opposite the largest angle.

An oblique triangle pairs side a with alpha, side b with beta, and side c with gamma. — The source labels the angles $\alpha,\beta,\gamma$ . These correspond to the angle measures $A,B,C$ used in this article.
Law of sines (simple).svg · Mliu92 · CC BY-SA 4.0.

Choose a rule from the known information

With ASA or AAS, two angles determine the third, and the sine rule determines missing sides. With SAS, two sides and their included angle determine the third side using the cosine rule. With SSS, the cosine rule determines an angle from all three sides.

SSA gives two sides and an angle not between them. Unlike SAS, it does not always determine a unique triangle. Work through every possible second angle before accepting an answer.

Choosing a triangle-solving method

Choosing a triangle-solving method
Known information	Start with	Important check
Two angles and one side	Angle sum, then sine rule	Third angle must be positive
Two sides and included angle	Cosine rule	Use the angle between the two known sides
Three sides	Cosine rule	Triangle inequality
Two sides and non-included angle	Sine rule, including both angle candidates	Zero, one or two triangles

Knowing only the three angles fixes the shape, not the size. Without a side length, there are infinitely many similar triangles.

The sine rule

For any non-degenerate triangle,

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

One way to understand the rule is to draw an altitude. The same height can be expressed using a side from either of two right triangles. Equating those expressions relates the corresponding side lengths and sines. The result also holds when the altitude falls outside an obtuse triangle.

Use the two fractions containing your known side-angle pair and the quantity you need. Writing all three fractions is not required in every calculation.

Worked example: two angles and a side

Let $A=40^\circ$ , $B=65^\circ$ and $a=12$ cm. The third angle is $75^\circ$ . Then

b=\frac{12\sin65^\circ}{\sin40^\circ}\approx16.92\text{ cm}

c=\frac{12\sin75^\circ}{\sin40^\circ}\approx18.03\text{ cm}

The lengths have the expected order: $a<b<c$ matches $40^\circ<65^\circ<75^\circ$ . Keep the exact sine expressions until the final rounding.

The cosine rule

The cosine rule gives

c^2=a^2+b^2-2ab\cos C

The angle $C$ must be the angle between the two sides $a$ and $b$ . If $C=90^\circ$ , the cosine term is zero and the rule becomes the Pythagorean theorem. For an acute included angle the cosine term reduces $c^2$ ; for an obtuse included angle it increases it.

Worked example: two sides and the included angle

Let $a=5$ cm, $b=7$ cm and $C=60^\circ$ .

c^2=25+49-70\cos60^\circ=39

c=\sqrt{39}\text{ cm}\approx6.24\text{ cm}

Choose the positive square root because $c$ is a side length. To finish the triangle, use the cosine rule for another angle or the sine rule with a known opposite pair, then check the angle sum. Here the remaining angles are approximately $A=43.90^\circ$ and $B=76.10^\circ$ . Their sum with $60^\circ$ is $180^\circ$ , and the angle opposite 7 cm is the largest.

Finding an angle from three sides

Rearrange the cosine rule before applying inverse cosine:

\cos C=\frac{a^2+b^2-c^2}{2ab}

For $a=5$ , $b=7$ , and $c=8$ ,

\cos C=\frac{25+49-64}{70}=\frac17

C=\arccos(1/7)\approx81.79^\circ

Inverse cosine returns an angle in $[0^\circ,180^\circ]$ , matching the possible range for a triangle’s interior angle. Calculating the largest angle first can make the overall size and shape easier to check.

Before calculating any angle, check that each side is shorter than the sum of the other two. Lengths 2, 3 and 6 cannot form a triangle. A value outside $[-1,1]$ for the computed cosine is a warning to recheck the data, units and algebra.

The ambiguous SSA case

Suppose $A=30^\circ$ , $a=7$ cm and $b=10$ cm. The sine rule gives

\sin B=\frac{b\sin A}{a}=\frac57

A calculator supplies one candidate, but a positive sine value is shared by an acute and a supplementary obtuse angle:

B_1=\arcsin(5/7)\approx45.5847^\circ

B_2=180^\circ-B_1\approx134.4153^\circ

Both candidates work here because adding $A=30^\circ$ leaves a positive third angle. Find each third angle and each third side separately:

Two triangles from the same SSA data
Candidate	Angle B	Angle C	Side c
First	45.58°	104.42°	13.56 cm
Second	134.42°	15.58°	3.76 cm

For each row, $C=180^\circ-A-B$ and $c=7\sin C/\sin30^\circ=14\sin C$ . The two solutions have different angle measures and third-side lengths.

One fixed angle and two side lengths allow two positions for a triangle vertex. — The two positions arise because a side of fixed length can meet the base ray in two places.
Sine Law – Ambiguous Case.svg · TheOtherJesse · Public domain (copyright-holder release).

Deciding whether there are zero, one or two triangles

For an acute given angle $A$ , set $h=b\sin A$ , where $b$ is the known side adjacent to $A$ and $a$ is the known opposite side. Compare $a$ with this height and with $b$ .

SSA count when A is acute

SSA count when A is acute
Condition	Number of non-degenerate triangles
$a<h$	0
$a=h$	1, with a right angle opposite b
$h<a<b$	2
$a\ge b$	1

In the worked example, $h=10\sin30^\circ=5$ , so $5<7<10$ predicts two triangles. If $a$ had been 4, no triangle would be possible; if it had been 5, there would be one.

For a right or obtuse given angle $A$ , its opposite side must be longer than the other known side: $a>b$ . That gives one triangle. If $a\le b$ , there is no non-degenerate triangle. Alternatively, the direct candidate-angle test works in every case: reject sine values above one, test the principal and supplementary candidates, and retain only positive third angles.

Area and final checks

If two sides and their included angle are known, the area is

K=\frac12ab\sin C

For sides 5 and 7 enclosing $60^\circ$ , $K=35\sqrt3/4$ cm², about 15.16 cm². This comes from one half times base times perpendicular height, with height expressed through sine.

An area formula needs the included angle. Mixing a side pair with an unrelated angle can produce a plausible number without representing the triangle.

For every solved triangle, check positive side lengths, an angle sum of $180^\circ$ , the triangle inequality, and the correspondence between large sides and large opposite angles. With SSA, perform these checks for each candidate rather than discarding the second angle automatically.

Use these tools in Bearings, Heights and Distances. The right-triangle definitions remain available in Trigonometric Ratios.

Practice questions

Find $b$ when $A=30^\circ$ , $B=45^\circ$ and $a=10$ .
For acute $A$ , suppose $b\sin A=5$ , $b=10$ and $a=7$ . How many triangles are possible?
A triangle has sides 5 and 7 enclosing $60^\circ$ . Find its area.

Check your answers

$b=10\sin45^\circ/\sin30^\circ=10\sqrt2$ .
Two, because $h<a<b$ . Each angle candidate still needs a positive third angle.
$K=\tfrac12(5)(7)\sin60^\circ=35\sqrt3/4$ square units.

Sources and further reading

Further explanations of the definitions and conventions used here:

OpenStax, Precalculus 2e: The sine rule and SSA ambiguity.
OpenStax, Precalculus 2e: The cosine rule.
Wolfram Language: Trigonometric Functions. Used for checking exact values, expressions and solutions.