Math Core

In geometry you learned that SSA is not a congruence shortcut: two sides and a non-included angle don't always pin down one triangle. Trigonometry shows exactly what can happen. Given SSA, there might be no triangle, exactly one, or two different triangles, and the law of sines tells you which.

Why SSA is ambiguous

Suppose you know angle $A$ , the side $b$ next to it, and the side $a$ across from it. Draw angle $A$ with side $b$ along one ray, ending at vertex $C$ . Side $a$ starts at $C$ and must reach the other ray of angle $A$ to finish the triangle. Think of side $a$ as a gate hinged at $C$ that can swing.

With A = 40°, b = 12 and a = 9, side a can swing to meet the base at two points, B₁ and B₂.

The dashed segment is the height from $C$ to the base line:

h = b \sin A.

It is the shortest possible distance from $C$ to that line. Compare side $a$ with $h$ and with $b$ :

If $a$ is shorter than $h$ , the swinging side can't reach the base line. No triangle.
If $a = h$ exactly, it just touches the line at a right angle. One right triangle.
If $a$ is longer than $h$ but shorter than $b$ , it hits the line in two places, $B_1$ and $B_2$ , both on the correct side of $A$ . Two triangles.
If $a \ge b$ , one of the two crossing points lands on or behind $A$ , so only one crossing point works. One triangle.

Counting SSA triangles

Given $A$ , $a$ and $b$ , let $h = b \sin A$ .

Angle $A$	Condition	Number of triangles
acute	$a < h$	$0$
acute	$a = h$	$1$ (a right triangle)
acute	$h < a < b$	$2$
acute	$a \ge b$	$1$
right or obtuse	$a \le b$	$0$
right or obtuse	$a > b$	$1$

When $A$ is right or obtuse, it is the largest angle, so the side across from it must be the longest. That's why you need $a > b$ in that case.

The algebraic check

You don't have to memorize the table. The law of sines gives the same answer on its own:

Compute $\sin B = \dfrac{b \sin A}{a}$ .
If $\sin B > 1$ , there is no triangle, because no angle has a sine greater than $1$ .
If $\sin B = 1$ , then $B = 90^\circ$ : one right triangle.
If $\sin B < 1$ , there are two candidates: $B_1 = \sin^{-1}(\sin B)$ , which is acute, and $B_2 = 180^\circ - B_1$ , which is obtuse. Keep each candidate only if $A + B < 180^\circ$ , so that angle $C$ is positive.

Worked example: No triangle

Is there a triangle with $A = 30^\circ$ , $b = 10$ and $a = 4$ ?

The height is $h = 10 \sin 30^\circ = 5$ , and $a = 4$ is shorter, so side $a$ can't reach. Algebraically,

\sin B = \frac{10 \sin 30^\circ}{4} = \frac{5}{4} = 1.25,

which is impossible. No triangle exists.

Worked example: Two triangles

Solve the triangle with $A = 40^\circ$ , $b = 12$ and $a = 9$ . Round to the nearest tenth.

The height is $h = 12 \sin 40^\circ \approx 7.71$ . Since $7.71 < 9 < 12$ , expect two triangles.

\sin B = \frac{12 \sin 40^\circ}{9} \approx 0.8570, \qquad B_1 = \sin^{-1}(0.8570) \approx 59.0^\circ, \qquad B_2 \approx 180^\circ - 59.0^\circ = 121.0^\circ.

Both work, because $40^\circ + 59.0^\circ$ and $40^\circ + 121.0^\circ$ are both less than $180^\circ$ . Finish each triangle using the complete pair $\dfrac{9}{\sin 40^\circ}$ :

	$B$	$C = 180^\circ - A - B$	$c = \dfrac{9 \sin C}{\sin 40^\circ}$
Triangle 1	$59.0^\circ$	$81.0^\circ$	$13.8$
Triangle 2	$121.0^\circ$	$19.0^\circ$	$4.6$

Both triangles have the given parts, $A = 40^\circ$ , $b = 12$ and $a = 9$ , yet they have different shapes.

Common mistake

The most common mistake is stopping at the calculator's answer. $\sin^{-1}$ only returns angles up to $90^\circ$ , so it never shows you the obtuse candidate. In SSA problems, always test $180^\circ - B_1$ as well.

Worked example: One triangle

Solve the triangle with $A = 35^\circ$ , $b = 8$ and $a = 11$ .

Here $a \ge b$ , so expect exactly one triangle.

\sin B = \frac{8 \sin 35^\circ}{11} \approx 0.4171, \qquad B_1 \approx 24.7^\circ, \qquad B_2 \approx 155.3^\circ.

The obtuse candidate fails: $35^\circ + 155.3^\circ > 180^\circ$ . So $B \approx 24.7^\circ$ , $C \approx 180^\circ - 35^\circ - 24.7^\circ = 120.3^\circ$ , and

c = \frac{11 \sin 120.3^\circ}{\sin 35^\circ} \approx 16.6.

(With unrounded values, $c \approx 16.55$ , which also rounds to $16.6$ .)

Worked example: An obtuse angle

Is there a triangle with $A = 110^\circ$ , $a = 7$ and $b = 9$ ?

Angle $A$ is obtuse, so $a$ must be the longest side. But $a = 7$ is shorter than $b = 9$ . No triangle. Check: $\sin B = \dfrac{9 \sin 110^\circ}{7} \approx 1.208$ , which is greater than $1$ .

Tip

A quick first look: in the acute case, if the side across from the given angle is at least as long as the other given side ( $a \ge b$ ), there is exactly one triangle and no ambiguity to worry about.

Practice

Practice 1

How many triangles have $A = 50^\circ$ , $b = 10$ and $a = 6$ ?

Practice 2

How many triangles have $A = 28^\circ$ , $b = 15$ and $a = 9$ ?

Practice 3

How many triangles have $A = 62^\circ$ , $b = 7$ and $a = 10$ ?

Practice 4

In triangle $ABC$ , $A = 40^\circ$ , $b = 12$ and $a = 10$ . Find all possible values of angle $B$ , to the nearest tenth of a degree. Separate answers with a comma.

Separate answers with commas, e.g. 2, -5

Practice 5

In triangle $ABC$ , $A = 42^\circ$ , $a = 16$ and $b = 22$ . Find all possible lengths of side $c$ , to the nearest tenth. Separate answers with a comma.

Separate answers with commas, e.g. 2, -5

Practice 6

In triangle $ABC$ , $A = 105^\circ$ , $a = 20$ and $b = 12$ . Find angle $B$ to the nearest tenth of a degree.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

In triangle $ABC$ , $A = 30^\circ$ and $b = 10$ . Which value of $a$ gives exactly two triangles?