The Ambiguous Case of the Sine Rule

One sine, two angles

The sine of an angle is the height of the point at that angle on a circle of radius 1. The points at 30° and at 150° are reflections of each other in the y-axis, so they are at the same height: sin 30° = sin 150° = ½.

In general, sin (180° − B) = sin B. So when the sine rule gives sin B = ½, it does not say whether B is 30° or 150°. The inverse sine key on a calculator gives only 30°, the acute one. The other candidate is always 180° minus the calculator answer.

The points at 30° and 150° on the circle. The dashed line joining them is level, so their heights, the two sines, are equal.

Both candidates can close a triangle

Suppose that in a triangle with A = 20° and the side a = 6 facing it, the sine rule has given sin B = ½. Try each candidate.

If B = 30°, the third angle is C = 180° − 20° − 30° = 130°. That is a triangle.

If B = 150°, the third angle is C = 180° − 20° − 150° = 10°. That is a triangle too: long and thin, with a 10° angle at C. Both triangles have A = 20°, the same side a = 6, and the same side b, so the same data describe two different triangles.

With B = 30°, the third angle is 130°.

The same A, a and b with B = 150°: the third angle is 10°.

The check: room for a third angle

Each candidate for B must leave a third angle greater than 0°, because the three angles of a triangle add to 180°. So the test is whether A + B is less than 180°.

With A = 20°, both candidates pass: 20° + 30° = 50° and 20° + 150° = 170° are both less than 180°.

Now take A = 75°, again with sin B = ½. The acute candidate gives 75° + 30° = 105°, leaving C = 75°. The obtuse candidate gives 75° + 150° = 225°, which is already past 180° before any third angle is added. So B = 30° is the only answer.

The acute candidate always passes when a triangle exists at all. The question is only ever whether the obtuse candidate fits as well.

With A = 75°, only B = 30° fits, and the third angle is 75°. An angle of 150° would not leave room for one.

Why a side can swing to two places

The two triangles come from the way the data are given: an angle A, the side b next to it, and the side a facing it. The angle A and the side b fix two corners and the direction of the third side. The side a then hangs from the end of b, and its other end must land on that line.

Swing side a like a pendulum. If it is shorter than the perpendicular distance from the end of b to the line, which is b sin A, it never reaches the line, and there is no triangle. If it equals b sin A, it just touches, at a right angle, and there is one triangle.

If it is longer than b sin A but shorter than b, it cuts the line in two places, and there are two triangles. If it is at least as long as b, the second crossing falls behind the corner A, and only one triangle is left.

A = 35° and b = 10 are fixed, so b sin A = 5.74. At a = 7.5 the swinging side meets the base twice: two triangles. Shorten a below 5.74 and it misses the base; lengthen it to 10 or more and only one crossing is left.

Solving both triangles

In a triangle, A = 30°, a = 6 cm and b = 10 cm. The sine rule gives $\sin B=10\times \sin 30^{\circ }÷6=5/6=0.8333$sin B = 10 × sin 30° ÷ 6 = 5 / 6 = 0.8333, to 4 decimal places.

Test first. b sin A = 10 × ½ = 5, and a = 6 is longer than 5 but shorter than 10, so there are two triangles.

The calculator gives $B={\sin }^{-1}\left(\frac{5}{6}\right)=56.44^{\circ }$B = sin⁻¹(5/6) = 56.44°, to 2 decimal places, and the other candidate is 180° − 56.44° = 123.56°. In the first triangle C = 180° − 30° − 56.44° = 93.56° and c = 6 × sin 93.56° ÷ sin 30° = 11.98 cm. In the second, C = 180° − 30° − 123.56° = 26.44° and c = 6 × sin 26.44° ÷ sin 30° = 5.34 cm, each to 2 decimal places.

Both answers are correct. Unless the question gives more information, such as whether the angle B is acute or obtuse, a full answer gives both triangles.

When it cannot happen

Two triangles are possible only when the given angle is not between the two given sides. With two sides and the angle between them, the triangle is fixed, and the cosine rule finds the third side directly.

The cosine rule never gives this problem when it finds an angle. A cosine is positive for an acute angle and negative for an obtuse one, so the sign of cos C says which kind of angle C is, and there is one answer.

And if the given angle A is obtuse, the other two angles must both be acute, so the obtuse candidate for B is always rejected.

The usual mistakes

Stopping at the calculator answer. The inverse sine gives only the acute angle; its supplement has the same sine and may give a second triangle.

Expecting two triangles every time. The obtuse candidate fits only if A + B is less than 180°, and when the side facing A is the longer of the two given sides it never does.

Pairing x with 360° − x. That is the pairing for a cosine. The angle with the same sine is 180° − x; below the axis, at 360° − x, the sine has the opposite sign.