Sine and Cosine Rules

Right triangles come with backstage support: the hypotenuse takes its place and sin, cos, tan are on call. But the triangles of the real world — measuring a river, locating a boat, designing a roof truss — usually have no right angle at all. They are oblique triangles, and they come with two weapons of their own: the sine rule and the cosine rule. Both answer the same kind of question — given some sides and angles, compute all the rest.

The sine rule: sides meet their angles

In any triangle, each side faces one angle. The sine rule says: each side divided by the sine of its opposite angle gives the same value for all three sides:

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

The intuition is smooth: a bigger angle opens wider and pushes its side longer, so the biggest angle must face the longest side. The sine rule turns this pairing into an exact equation. It shines when you know two angles and any one side: the third angle comes free from the $180^\circ$ angle sum, and one more proportion delivers every remaining side.

Measuring a river

A tree stands at $T$ on the far bank. Lay out a baseline $AB = 100$ m along your bank. At $A$ , the sightline $AT$ makes $74^\circ$ with the bank; at $B$ , the sightline $BT$ makes $61^\circ$ . In triangle $ABT$ , the angle at the tree is $180 - 74 - 61 = 45^\circ$ . Side $AT$ faces angle $B$ , and $AB$ faces angle $T$ , so the sine rule gives:

$\frac{AT}{\sin 61^\circ} = \frac{100}{\sin 45^\circ}$

$AT = 100 \times \frac{\sin 61^\circ}{\sin 45^\circ} \approx 100 \times \frac{0.8746}{0.7071} \approx 124$

About 124 meters from $A$ to the tree — the river is measured without stepping into the water.

The cosine rule: Pythagoras, upgraded

The Pythagorean theorem serves right angles only. Replace the right angle with any angle $C$ , and the equation takes a discount:

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

When $C = 90^\circ$ , $\cos C = 0$ and the whole subtracted term vanishes, returning the Pythagorean theorem — that is why this is a genuine extension. The smaller the angle $C$ , the closer $\cos C$ is to 1, the more gets subtracted, and the shorter $c$ becomes; the bigger the angle, the longer $c$ . The rule handles two jobs well: given two sides and the included angle, find the third side; given all three sides, reverse it ( $\cos C = \frac{a^2+b^2-c^2}{2ab}$ ) to find an angle. Reading it backward works just as smoothly — with three sides known, compute the largest angle first, then hand the remaining two to the sine rule, far more precise than any protractor.

Checking with 3-4-5

Take the familiar 3-4-5 right triangle: $a = 3$ , $b = 4$ , $c = 5$ , $C = 90^\circ$ . Substitute into the cosine rule: $25 = 9 + 16 - 2 \times 3 \times 4 \times 0 = 25$ , exactly right — the rule has degenerated into Pythagoras.

Now tilt the angle: with the same 3 and 4 but an included angle of $C = 60^\circ$ , where $\cos 60^\circ = 0.5$ , we get $c^2 = 9 + 16 - 24 \times 0.5 = 13$ , so $c = \sqrt{13} \approx 3.61$ . Shrinking the angle from $90^\circ$ to $60^\circ$ shortened the third side, as promised.

Which weapon, when

Match the tool to what you know:

Two angles and any side → the sine rule (third angle included free);
Two sides and the angle between them → the cosine rule finds the third side;
All three sides → the rearranged cosine rule finds an angle;
A right angle → the old friends Pythagoras plus trigonometry are enough.

Mnemonic: look for the included angle. If it is there, use the cosine rule; if the given angles are ones whose opposite sides you know, use the sine rule. To push ratios and identities further, visit Trig Identities; the classification and properties of triangles live in Triangles.

Check yourself

Quick quiz

1. Which equation is the sine rule?
2. With a=3, b=4 and C=90°, the cosine rule gives c = ?
3. You know two sides and the angle between them. Which weapon first?