Finding Angles with the Sine Rule

The rule runs both ways

The sine rule says that in any triangle $a/\sin A=b/\sin B=c/\sin C$a / sin A = b / sin B = c / sin C: each side divided by the sine of the angle facing it gives the same number. Any one equation from it, such as $a/\sin A=b/\sin B$a / sin A = b / sin B, holds two sides and two angles.

Used to find a side, it needs two angles and one side. But the equation does not care which of its four parts is missing. Given two sides and the angle facing one of them, it finds the sine of the angle facing the other side, and from the sine, the angle.

Three parts known, one to find

In a triangle, the side of 8 cm faces an angle of 30°, and the side of 16 cm faces the angle B. The pair 8 and 30° is complete. The side 16 is known but the angle B facing it is not.

So in the equation $8/\sin 30^{\circ }=16/\sin B$8 / sin 30° = 16 / sin B, three of the four parts are known, and the only unknown is sin B.

The side 8 faces the angle of 30°, and the side 16 faces the unknown angle B.

Rearrange for the sine

When the unknown is an angle, it is easier to write the rule upside down, with the sines on top: $\sin A/a=\sin B/b$sin A / a = sin B / b. Both forms are true, because if two fractions are equal, their reciprocals are equal too.

Here that reads $\sin 30^{\circ }÷8=\sin B/16$sin 30° ÷ 8 = sin B / 16. Multiply both sides by 16: sin B = 16 × sin 30° ÷ 8.

Now work it out. sin 30° = ½, so $\sin B=16\times \frac{1}{2}÷8=8/8=1$sin B = 16 × ½ ÷ 8 = 8 / 8 = 1.

From the sine to the angle

The only angle between 0° and 180° with a sine of 1 is 90°. So B = 90°, and the triangle has a right angle at B.

That fits: in a right triangle, the side facing 30° is half the hypotenuse, and 8 is half of 16.

The exact values hand back their angles in the same way. A sine of ½ comes from 30°, a sine of $\frac{1}{\sqrt{2}}$1/√2 from 45°, and a sine of $\frac{\sqrt{3}}{2}$√3/2 from 60°. For any other value, the inverse sine key on a calculator gives the angle.

The sine of the angle is the height of the point on a circle of radius 1. Turn the arm: the height grows to 1 at 90°, at the top of the circle, and nowhere else.

A calculator angle, and a second candidate

In a triangle, a = 10 cm, A = 50° and b = 7 cm. Then $\sin B/7=\sin 50^{\circ }÷10$sin B / 7 = sin 50° ÷ 10, so sin B = 7 × sin 50° ÷ 10. With sin 50° = 0.7660, to 4 decimal places, $\sin B=5.362/10=0.5362$sin B = 5.362 / 10 = 0.5362.

The inverse sine gives $B={\sin }^{-1}\left(0.5362\right)=32.43^{\circ }$B = sin⁻¹(0.5362) = 32.43°, to 2 decimal places. But an angle and its supplement have the same sine, so 180° − 32.43° = 147.57° has a sine of 0.5362 too. Every angle found from a sine has this second candidate, and it has to be checked.

Here it fails. The angles of a triangle add to 180°, and 50° + 147.57° = 197.57° is already more than 180°. So B = 32.43°, and the third angle is C = 180° − 50° − 32.43° = 97.57°.

There is a quicker test. The side b = 7 is shorter than a = 10, so the angle B facing it must be smaller than A = 50°. An obtuse B is impossible.

a = 10 cm faces 50°, and b = 7 cm faces B. The shorter side faces the smaller angle, so B = 32.43° and not its supplement.

When both candidates fit

Change the triangle so that the known angle faces the shorter side: A = 30°, a = 8 cm and b = 12 cm. Then $\sin B=12\times \sin 30^{\circ }÷8=6/8=0.75$sin B = 12 × sin 30° ÷ 8 = 6 / 8 = 0.75, and ${\sin }^{-1}\left(0.75\right)=48.59^{\circ }$sin⁻¹(0.75) = 48.59°, to 2 decimal places.

The second candidate is 180° − 48.59° = 131.41°, and 30° + 131.41° = 161.41° leaves 18.59° for the third angle. Both values of B make a triangle with these measurements.

So check the supplement every time. When the known angle faces the longer of the two known sides, the supplement never fits and there is one answer. When it faces the shorter side, there can be two.

The usual mistakes

Pairing a side with the wrong angle. Each side goes with the angle facing it. In this triangle, 16 goes with B, not with the 30°.

Taking the inverse sine of the side ratio. sin B is 16 × sin 30° ÷ 8, not 16 ÷ 8; the sine of the known angle is part of the calculation.

Stopping at the calculator answer. The inverse sine gives only the acute angle. Check 180° minus it, and keep it only if it leaves room for a third angle.

Repeating the given angle. A side twice as long as another faces a much larger angle, so B cannot be 30° here.

An angle after the cosine rule

The application below first finds a side with the cosine rule: with two sides and the angle between them known, the third side squared is the sum of the two squares minus twice their product times the cosine of that angle.

Once all three sides are known, the angle at J is found with the sine rule, exactly as above. The calculator gives an acute angle, and its supplement is rejected because the side facing J is shorter than the side facing the 72° angle.