Grade 11 · MCR3U · Functions

MCR3U: Sine Law and the Ambiguous Case

The sine law solves any triangle where you know a side and the angle across from it. One setup, two sides and an angle that is not between them, can give zero, one or two triangles. That is the ambiguous case.

How It Works

The sine law. In any triangle, each side divided by the sine of its opposite angle gives the same number: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ . Use it when you know a matching pair, a side and the angle across from it. Without a matching pair, start with the cosine law instead. Put the unknown in the numerator: write $\frac{b}{\sin B} = \frac{a}{\sin A}$ to find a side and $\frac{\sin B}{b} = \frac{\sin A}{a}$ to find an angle.

Why it gets ambiguous. When you solve for an angle, you get $\sin B$ , and two angles between $0^\circ$ and $180^\circ$ share every sine value below $1$ : $B$ and $180^\circ - B$ . Your calculator only shows the acute one. With two angles and a side there is no problem, because the third angle is forced. With two sides and a non-included angle (SSA), both answers can be real.

The test before you solve. Say you know angle $A$ , the side $a$ across from it, and a second side $b$ . Swing side $a$ from the end of $b$ like a gate. The shortest it could be and still reach the base is the height $h = b\sin A$ . For an acute $A$ :

If $a < h$ , side $a$ is too short to reach. No triangle.

If $a = h$ , it just touches. One right triangle.

If $h < a < b$ , it reaches the base in two places. Two triangles.

If $a \ge b$ , the second spot falls on the wrong side of $A$ . One triangle.

If $A$ is obtuse, there is one triangle when $a > b$ and none otherwise.

The test while you solve. Find $\sin B$ . If it is greater than $1$ , there is no triangle. Otherwise take $B$ from your calculator and also try $180^\circ - B$ . The second one is a real triangle only if it leaves room for $C$ , meaning $A + B < 180^\circ$ . Then finish each triangle separately.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
In $\triangle ABC$ , $\angle A = 40^\circ$ , $\angle B = 75^\circ$ and $a = 12$ . Find $b$ and $c$ .
Show solution
1. $\angle C = 180^\circ - 40^\circ - 75^\circ = 65^\circ$ . Two angles are known, so this is not ambiguous.
2. $b = \frac{12\sin 75^\circ}{\sin 40^\circ} \approx 18.03$ .
3. $c = \frac{12\sin 65^\circ}{\sin 40^\circ} \approx 16.92$ .
Answer: $\angle C = 65^\circ$ , $b \approx 18.03$ , $c \approx 16.92$
Easy
$\angle A = 50^\circ$ , $a = 6$ and $b = 10$ . How many triangles are there?
Show solution
1. Height: $h = 10\sin 50^\circ \approx 7.66$ . Since $a = 6 < 7.66$ , side $a$ cannot reach.
2. The sine law agrees: $\sin B = \frac{10\sin 50^\circ}{6} \approx 1.277$ , and no angle has a sine greater than $1$ .
Answer: No triangle.
Medium
$\angle A = 35^\circ$ , $a = 14$ and $b = 10$ . Solve the triangle.
Show solution
1. $a \ge b$ , so expect one triangle.
2. $\sin B = \frac{10\sin 35^\circ}{14} \approx 0.4097$ , so $B \approx 24.2^\circ$ .
3. Try the other angle: $180^\circ - 24.2^\circ = 155.8^\circ$ , but $35^\circ + 155.8^\circ > 180^\circ$ . Rejected.
4. $\angle C \approx 180^\circ - 35^\circ - 24.2^\circ = 120.8^\circ$ and $c = \frac{14\sin 120.8^\circ}{\sin 35^\circ} \approx 20.96$ .
Answer: One triangle: $\angle B \approx 24.2^\circ$ , $\angle C \approx 120.8^\circ$ , $c \approx 20.96$
Medium
$\angle A = 32^\circ$ , $a = 9$ and $b = 15$ . Solve every possible triangle.
Show solution
1. Height: $h = 15\sin 32^\circ \approx 7.95$ . Since $7.95 < 9 < 15$ , expect two triangles.
2. $\sin B = \frac{15\sin 32^\circ}{9} \approx 0.8832$ , so $B \approx 62.0^\circ$ or $B \approx 180^\circ - 62.0^\circ = 118.0^\circ$ . Both leave room: $32^\circ + 118.0^\circ < 180^\circ$ .
3. Triangle 1: $\angle C \approx 86.0^\circ$ and $c = \frac{9\sin 86.0^\circ}{\sin 32^\circ} \approx 16.94$ .
4. Triangle 2: $\angle C \approx 30.0^\circ$ and $c = \frac{9\sin 30.0^\circ}{\sin 32^\circ} \approx 8.50$ .
Answer: Two triangles: $\angle B \approx 62.0^\circ$ , $\angle C \approx 86.0^\circ$ , $c \approx 16.94$ ; or $\angle B \approx 118.0^\circ$ , $\angle C \approx 30.0^\circ$ , $c \approx 8.50$
Test level
In $\triangle ABC$ , $\angle A = 40^\circ$ and $b = 12$ . For which lengths of $a$ are there no triangles, one triangle, or two?
Show solution
1. Height: $h = 12\sin 40^\circ \approx 7.71$ .
2. $a < 7.71$ : too short to reach. No triangle.
3. $a = 12\sin 40^\circ \approx 7.71$ : one right triangle, with $\angle B = 90^\circ$ .
4. $7.71 < a < 12$ : two triangles.
5. $a \ge 12$ : one triangle.
Answer: None for $a < 7.71$ ; one for $a \approx 7.71$ or $a \ge 12$ ; two for $7.71 < a < 12$ (to two decimals).

Learn It on Martian Lab

Martian Lab teaches this in the Solving Triangles & the Ambiguous Case lesson of its MCR3U Trigonometric Ratios unit. Watch it, then take the Trigonometric Ratios practice tests to see if it stuck.

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