Sine and Cosine Rules

Trigonometry connects the angles of a triangle to the lengths of its sides. Three ratios named sine, cosine, and tangent do most of the work.

Sine and Cosine Rules: the central idea

The sine rule uses known opposite pairs; the cosine rule connects three sides with an included angle and contains Pythagoras as the $90^\circ$ case.

Words you need

Sine rule

The sine rule states that $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ in every triangle, meaning each side divided by the sine of its opposite angle gives the same number.

Cosine rule

The cosine rule states that $c^2=a^2+b^2-2ab\cos C$, linking all three sides to one angle, and it is the tool for triangles where no side-and-opposite-angle pair is known.

Included angle

The included angle is the angle wedged between the two sides you were given, and it is what makes a set of measurements SAS rather than SSA.

Opposite side

The opposite side is the side lying directly across from an angle and never touching it, which is why side $a$ is always paired with angle $A$.

Ambiguous case

The ambiguous case is the SSA situation, where two sides and a non-included angle can describe two different triangles, one triangle, or none at all, so the supplementary angle must always be tested.

Circumradius

The circumradius $R$ is the radius of the circle passing through all three corners of a triangle, and the sine rule extends to $\dfrac{a}{\sin A}=2R$, so the shared ratio is not an accident.

What to know before this lesson

Know triangle notation, opposite side-angle pairs, inverse trig functions, and the $180^\circ$ angle sum.

If one of those prerequisites is uncertain, use the Trigonometry subject guide to locate the earlier concept before memorizing a procedure.

The sine and cosine rules: a worked example

Every step, with the arithmetic

1. Step 1 - Label everythingA triangle has $b=7$, $c=10$, and the angle between them $A=60^\circ$. Side $a$, across from $A$, is unknown. This is the SAS case.
2. Step 2 - Pick the ruleNo side is paired with its opposite angle yet, so the sine rule cannot start. Use $a^2=b^2+c^2-2bc\cos A$.
3. Step 3 - Substitute the numbers$a^2=7^2+10^2-2(7)(10)\cos60^\circ$
4. Step 4 - Do the squares and the cosine$a^2=49+100-140(0.5)=149-70=79$
5. Step 5 - Take the square root$a=\sqrt{79}\approx8.89$
6. Step 6 - Now the sine rule can find a second angle$\sin B=\dfrac{b\sin A}{a}=\dfrac{7(0.8660)}{8.89}=\dfrac{6.062}{8.89}=0.6821$, so $B\approx43.0^\circ$.
7. Step 7 - Subtract for the last angle$C=180^\circ-60^\circ-43.0^\circ=77.0^\circ$
8. Step 8 - Check it makes senseThe longest side is $c=10$ and the largest angle is $C=77.0^\circ$, sitting directly across from it. The shortest side $b=7$ faces the smallest angle. Consistent.

In $c^2=a^2+b^2-2ab\cos C$, setting $C=90^\circ$ makes $\cos C=0$, leaving $c^2=a^2+b^2$.

Which rule solves which triangle, and what can go wrong

The step-by-step method for the sine and cosine rules

1. Label sides $a,b,c$ opposite angles $A,B,C$ before selecting a formula.
2. Use the sine rule for a known opposite pair and the cosine rule for SAS or SSS information.
3. Check the ambiguous SSA case, angle sum, and whether every computed side satisfies the triangle inequality.

How to check your answer

Verify all three angles add to $180^\circ$ and the largest angle lies opposite the longest side.

Confirm the quadrant and sign, convert degrees and radians consistently, and substitute solutions into the original interval because periodic equations usually have more than one answer.

A mistake that changes the mathematics

With the sine rule, $\sin^{-1}$ returns one principal angle; an SSA problem may also admit the supplementary angle.

Where you will actually use this

Surveying across an obstacle

To measure across a river without crossing it, a surveyor marks a $200$-metre baseline on one bank and sights a tree on the far bank from both ends, reading angles of $65^\circ$ and $78^\circ$. The third angle is $37^\circ$, and the sine rule gives the distance to the tree as $200\sin78^\circ/\sin37^\circ\approx325$ metres.

Air and sea navigation

A plane flies $120$ km, turns $40^\circ$, then flies another $90$ km. The interior angle of the triangle is $180^\circ-40^\circ=140^\circ$, so the direct distance home is $\sqrt{120^2+90^2-2(120)(90)\cos140^\circ}\approx\sqrt{39046}\approx198$ km.

Forces and truss design

When two cables pull on a joint at a known angle, the resulting force is the third side of a triangle, found with the cosine rule. Engineers use this to check that no member of a roof truss or crane carries more load than it was rated for.

How the sine and cosine rules connects to the rest of trigonometry

• Pythagorean theorem — Pythagoras is the cosine rule's right-angle special case, so learning the cosine rule extends a tool you already trust to triangles with no square corner.
• Area of a triangle — The area formula $A=\tfrac12ab\sin C$ uses exactly the same side and angle labels, so once a triangle is solved its area is one more line of arithmetic.
• The unit circle — The unit circle explains why $\cos140^\circ$ is negative, which is what makes the cosine rule add rather than subtract when the angle is obtuse.

Try a transfer problem

Given $a=8$, $b=11$, and $A=35^\circ$, determine whether zero, one, or two triangles are possible.

Show the worked answer

With $a=8$, $b=11$, and $A=35^\circ$, this is SSA, so check the ambiguous case. The height of the triangle from the far corner down to the base is $h=b\sin A=11\sin35^\circ=11(0.5736)\approx6.31$. Compare: $h\approx6.31<a=8<b=11$. When the given side is longer than the height but shorter than the other given side, two triangles fit, so the answer is two. Working them out, the sine rule gives $\sin B=\dfrac{b\sin A}{a}=\dfrac{6.31}{8}=0.7887$, so $B\approx52.1^\circ$ or its supplement $B\approx127.9^\circ$. Both survive, because $35^\circ+52.1^\circ=87.1^\circ$ and $35^\circ+127.9^\circ=162.9^\circ$ are each below $180^\circ$. The first triangle has $C\approx92.9^\circ$ and $c=\dfrac{8\sin92.9^\circ}{\sin35^\circ}\approx13.9$. The second has $C\approx17.1^\circ$ and $c=\dfrac{8\sin17.1^\circ}{\sin35^\circ}\approx4.1$.

Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to trigonometry practice questions.

Questions about the sine and cosine rules

How do I decide between the sine rule and the cosine rule?

Look for a matched pair: a side together with the angle straight across from it. If you have one, the sine rule works. If you do not, you were given SAS or SSS, and the cosine rule is the only way in. After the cosine rule produces a missing side, a matched pair exists and you can switch to the sine rule for the rest.

Why can SSA give two triangles?

Because $\sin B$ and $\sin(180^\circ-B)$ are equal, so a sine value of $0.7887$ points at both $52.1^\circ$ and $127.9^\circ$. Your calculator only reports the first. Test whether the supplement still leaves room under $180^\circ$ once the known angle is added; if it does, a second triangle genuinely exists.

What does a negative cosine in my answer mean?

It means the angle is obtuse, somewhere between $90^\circ$ and $180^\circ$. If $\cos A=-0.5$ then $A=120^\circ$. This is a strength of the cosine rule over the sine rule: cosine gives one unambiguous angle in a triangle, so it never suffers the two-answer problem.

Can the cosine rule find an angle instead of a side?

Yes, rearranged as $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$. For a triangle with sides $5$, $6$, and $7$, finding the angle across from the $5$: $\cos A=\dfrac{36+49-25}{2(6)(7)}=\dfrac{60}{84}=0.7143$, so $A\approx44.4^\circ$. This is the standard first move for SSS.