The Ambiguous Case

Usually, three pieces of information pin down exactly one triangle. But if you know two sides and an angle that isn’t between them (SSA), there might be no triangle, one triangle, or two different triangles that fit. This is called the ambiguous case of the sine law. All angles are in degrees.

Key ideas

Why there can be two triangles

Suppose you know $\angle A$, the side $b$ next to it, and the side $a$ opposite it. Picture side $a$ hanging from $C$ and swinging like a pendulum. It might miss the base entirely, touch it once, or hit it at two points.

With $\angle A = 40^\circ$, $b = 12$, and $a = 9$, side $a$ can land at $B_1$ or $B_2$.

Counting the triangles (when angle A is acute)

First find the height from $C$ to the base: $h = b\sin A$. Then compare $a$ with $h$ and $b$:

Condition	Number of triangles
$a \lt h$	none (side $a$ is too short to reach)
$a = h$	one (a right triangle)
$h \lt a \lt b$	two
$a \ge b$	one

If $\angle A$ is obtuse, there’s one triangle when $a \gt b$, and none when $a \le b$.

Solving with the sine law

1. Use the sine law to find $\sin B$.
2. If $\sin B \gt 1$, there’s no triangle.
3. Otherwise, $B_1 = \sin^{-1}(\dots)$ and $B_2 = 180^\circ - B_1$.
4. Keep $B_2$ only if $\angle A + B_2 \lt 180^\circ$.
5. Finish each triangle: find $\angle C$, then side $c$.

Worked examples

Example 1: Two triangles

In $\triangle ABC$, $\angle A = 40^\circ$, $b = 12$, and $a = 9$. Solve all possible triangles.

Solution. $h = 12\sin 40^\circ \approx 7.71$. Since $7.71 \lt 9 \lt 12$, there are two triangles.

$$\sin B = \frac{12\sin 40^\circ}{9} \approx 0.8570$$

• Triangle 1: $B_1 \approx 59.0^\circ$, $C_1 \approx 180^\circ - 40^\circ - 59.0^\circ = 81.0^\circ$, and $c_1 = \dfrac{9\sin C_1}{\sin 40^\circ} \approx 13.83$.
• Triangle 2: $B_2 \approx 180^\circ - 59.0^\circ = 121.0^\circ$, $C_2 \approx 19.0^\circ$, and $c_2 = \dfrac{9\sin C_2}{\sin 40^\circ} \approx 4.56$.

Example 2: No triangle

In $\triangle ABC$, $\angle A = 50^\circ$, $b = 10$, and $a = 6$. How many triangles are there?

Solution. $h = 10\sin 50^\circ \approx 7.66$, and $a = 6$ is shorter than that, so no triangle exists.

The sine law agrees: $\sin B = \dfrac{10\sin 50^\circ}{6} \approx 1.277$, which is impossible, since sine is never more than $1$.

Example 3: One triangle

In $\triangle ABC$, $\angle A = 35^\circ$, $b = 8$, and $a = 10$. Solve the triangle.

Solution. $a \ge b$, so there’s exactly one triangle.

$$\sin B = \frac{8\sin 35^\circ}{10} \approx 0.4589 \quad\Rightarrow\quad B \approx 27.3^\circ$$

The other option, $180^\circ - 27.3^\circ = 152.7^\circ$, doesn’t fit, because $35^\circ + 152.7^\circ \gt 180^\circ$.

$\angle C \approx 180^\circ - 35^\circ - 27.3^\circ = 117.7^\circ$, and $c = \dfrac{10\sin C}{\sin 35^\circ} \approx 15.44$.

Example 4: Exactly one right triangle

In $\triangle ABC$, $\angle A = 30^\circ$, $b = 10$, and $a = 5$. Describe the triangle.

Solution. $h = 10\sin 30^\circ = 5 = a$. Side $a$ just reaches the base, so there’s one triangle, with a right angle at $B$.

Check: $\sin B = \dfrac{10\sin 30^\circ}{5} = 1$, so $\angle B = 90^\circ$.

Common mistakes

Missing the second triangle. $\sin^{-1}$ only gives the acute angle. Always check whether $180^\circ - B$ also fits.

Keeping an impossible second angle. If $\angle A + B_2 \ge 180^\circ$, the second triangle doesn’t exist.

Using the wrong height. $h = b\sin A$ uses the side next to the known angle ($b$), not the side opposite it.

Applying the ambiguous case to other information. Only SSA can be ambiguous. Two angles and a side, or SAS, or SSS, always give at most one triangle.

Rounding $B_1$ before finding $B_2$. Keep full values so the second triangle’s numbers aren’t off.

Practice

1. (Warm-up) For $\angle A = 30^\circ$, $b = 20$, and $a = 12$, find $h$ and say how many triangles there are.

Solution

$h = 20\sin 30^\circ = 10$. Since $10 \lt 12 \lt 20$, there are two triangles.

2. (Warm-up) How many triangles have $\angle A = 60^\circ$, $b = 10$, and $a = 12$?

Solution

One, because $a \ge b$.

3. (Warm-up) How many triangles have $\angle A = 45^\circ$, $b = 10$, and $a = 6$?

Solution

$h = 10\sin 45^\circ \approx 7.07$. Since $a = 6 \lt h$, there’s no triangle.

4. (Core) Solve all triangles with $\angle A = 32^\circ$, $b = 15$, and $a = 10$.

Solution

$h = 15\sin 32^\circ \approx 7.95$, and $7.95 \lt 10 \lt 15$, so two triangles.

$\sin B = \dfrac{15\sin 32^\circ}{10} \approx 0.7949$.

• Triangle 1: $B_1 \approx 52.6^\circ$, $C_1 \approx 95.4^\circ$, $c_1 \approx 18.79$.
• Triangle 2: $B_2 \approx 127.4^\circ$, $C_2 \approx 20.6^\circ$, $c_2 \approx 6.65$.

5. (Core) How many triangles have $\angle A = 120^\circ$, $a = 9$, and $b = 11$?

Solution

$\angle A$ is obtuse and $a \le b$, so there’s no triangle. (The side opposite an obtuse angle must be the longest side.)

6. (Core) Solve the triangle with $\angle A = 25^\circ$, $a = 14$, and $b = 9$.

Solution

$a \ge b$, so one triangle.

$$\sin B = \frac{9\sin 25^\circ}{14} \approx 0.2717 \quad\Rightarrow\quad B \approx 15.8^\circ$$

$\angle C \approx 139.2^\circ$ and $c = \dfrac{14\sin C}{\sin 25^\circ} \approx 21.63$.

7. (Core) In $\triangle ABC$, $\angle A = 40^\circ$ and $b = 12$. For which lengths of $a$ are there two triangles?

Solution

Two triangles need $h \lt a \lt b$, where $h = 12\sin 40^\circ \approx 7.71$. So $7.71 \lt a \lt 12$ (more precisely, $12\sin 40^\circ \lt a \lt 12$).

8. (Challenge) A straight road runs past a cell tower. From point $A$ on the road, the tower is $10$ km away, at an angle of $30^\circ$ to the road. The tower’s signal reaches $7$ km. Between which distances along the road from $A$ (in the direction of the tower) does a car get a signal?

Solution

This is SSA with $\angle A = 30^\circ$, $b = 10$, and $a = 7$. Since $h = 10\sin 30^\circ = 5 \lt 7 \lt 10$, there are two points on the road exactly $7$ km from the tower.

Let $x$ be the distance along the road. By the cosine law:

$$7^2 = 10^2 + x^2 - 2(10)x\cos 30^\circ \quad\Rightarrow\quad x^2 - 10\sqrt{3}\,x + 51 = 0$$

The quadratic formula gives $x \approx 3.76$ or $x \approx 13.56$. The car has a signal between about $3.76$ km and $13.56$ km along the road.