The Sine Rule

Sides and the angles facing them

Name the corners of a triangle A, B and C, and the angles at them A, B and C as well. Each side is named with the small letter of the corner across from it: side a is opposite the corner A, side b is opposite B, and side c is opposite C.

So every side has one angle facing it from across the triangle, and that angle carries the same letter. Side a and angle A are a pair, and so are b and B, and c and C. The sine rule is a statement about these three pairs.

Each side points across the triangle to the angle facing it: a to A, b to B, and c to C.

One height, two right triangles

The triangle has no right angle, so the ratios of a right triangle cannot be used on it directly. Make right triangles: draw the height h from the corner A, perpendicular to side a. It splits the triangle into two right triangles that share the side h.

In the right triangle on the side of B, the hypotenuse is c and h is opposite the angle B. So $\sin B=h/c$sin B = h / c, which gives h = c sin B.

In the right triangle on the side of C, the hypotenuse is b and h is opposite the angle C. So $\sin C=h/b$sin C = h / b, which gives h = b sin C.

Both are the same length h, so c sin B = b sin C.

The height h from A splits the triangle into two right triangles. In one, h = c sin B; in the other, h = b sin C.

Each side over the sine of its angle

Divide both sides of c sin B = b sin C by sin B sin C. On the left the sin B cancels, leaving $c/\sin C$c / sin C. On the right the sin C cancels, leaving $b/\sin B$b / sin B. So $c/\sin C=b/\sin B$c / sin C = b / sin B.

Each side now sits over the sine of the angle facing it. Dropping the height from B instead, onto side b, gives c sin A = a sin C by the same steps, and so $a/\sin A=c/\sin C$a / sin A = c / sin C.

Put the two results together: $a/\sin A=b/\sin B=c/\sin C$a / sin A = b / sin B = c / sin C. This is the sine rule. In any triangle, each side divided by the sine of the angle facing it gives the same number.

When the rule can be used

Each equation in the sine rule links two pairs, so it has four quantities in it: two sides and the two angles facing them. Know three of them and the fourth can be found.

For finding a side, that means knowing one complete pair, a side and the angle facing it, plus one more angle. If two angles are given, the third is 180° minus their sum, so knowing any two angles and any one side is enough.

If the information is two sides and the angle between them, or three sides, no complete pair is known. The sine rule has two unknowns in every equation, and the cosine rule is needed instead.

Side a = 8 cm and the angle A = 40° facing it are a complete pair, and the angle B = 75° faces the unknown side b.

Solving the triangle

In the triangle above, a = 8 cm, A = 40° and B = 75°. Use the pair a and A with the angle B: $b/\sin 75^{\circ }=8/\sin 40^{\circ }$b / sin 75° = 8 / sin 40°.

Multiply both sides by sin 75°: b = 8 × sin 75° ÷ sin 40°. With sin 75° = 0.9659 and sin 40° = 0.6428, to 4 decimal places, $b=7.727/0.6428=12.02$b = 7.727 / 0.6428 = 12.02 cm, to 2 decimal places.

The third angle is C = 180° − 40° − 75° = 65°. Then $c=8\times \sin 65^{\circ }÷\sin 40^{\circ }=8\times 0.9063/0.6428=11.28$c = 8 × sin 65° ÷ sin 40° = 8 × 0.9063 / 0.6428 = 11.28 cm, to 2 decimal places.

Check the order. The largest angle, 75°, faces the longest side, 12.02 cm. The smallest angle, 40°, faces the shortest side, 8 cm. In every triangle the longest side faces the largest angle, so a result out of that order means a pairing has gone wrong.

An exact answer

In a triangle with A = 30°, a = 5 and B = 45°, the common value of the rule is $5/\sin 30^{\circ }=5÷\frac{1}{2}=10$5 / sin 30° = 5 ÷ ½ = 10. So every side is 10 times the sine of the angle facing it.

That gives $b=10\sin 45^{\circ }=10\times \frac{\sqrt{2}}{2}=5\sqrt{2}$b = 10 sin 45° = 10 × √2/2 = 5√2, which is 7.07 to 2 decimal places. The angle C = 180° − 30° − 45° = 105°, and c = 10 sin 105° = 9.66, to 2 decimal places, the longest side, facing the largest angle.

When one angle is obtuse

If the angle B is obtuse, the height from A falls outside the triangle, onto side a extended past B. The right triangle on that side then contains the angle 180° − B, so h = c sin (180° − B).

An angle and its supplement have the same sine, so sin (180° − B) = sin B and h = c sin B after all. The rest of the argument is unchanged, and the sine rule holds in every triangle, obtuse or not.

The usual mistakes

Pairing a side with the wrong angle. In the sine rule a side goes with the angle facing it, across the triangle, never with an angle at one of its own ends.

Turning one fraction upside down. Write the rule with the unknown on top: $b/\sin B=a/\sin A$b / sin B = a / sin A, then multiply both sides by sin B.

Multiplying by sin A instead of dividing by it. In $b=a\times \sin B/\sin A$b = a × sin B / sin A, the sine of the known angle is on the bottom.

Using the rule without a complete pair. With two sides and the angle between them, every equation in the sine rule has two unknowns; that triangle needs the cosine rule.