Math · Precalculus · Worked example

Find both triangles in the ambiguous case

In triangle ABC, $A = 30 °$ , $a = 5$ and $b = 8$ . Find every triangle that fits.

Recognize SSA

The known angle $A$ is opposite side $a$ , not between the two known sides. That is SSA, the one case that can have two answers.

Find $sin B$

Side $a$ and angle $A$ are a complete pair, so the law of sines gives the sine of the angle opposite $b$ . Since $sin 30 ° = 1 / 2$ , $sin B = 8 (1 / 2) / 5 = 4 / 5$ .

\begin{array}{l} \frac{sin B}{8} = \frac{sin 30^{\circ}}{5} \\ sin B = \frac{8 sin 30^{\circ}}{5} \\ sin B = \frac{4}{5} \end{array}

List both candidates

$0.8$ is less than $1$ , so there are angles with this sine. The inverse sine gives $B \approx 53.13 °$ , and its supplement, $180 ° - 53.13 ° = 126.87 °$ , has the same sine.

B \approx {53.13}^{\circ} or B \approx {126.87}^{\circ}

Test each candidate

$30 ° + 53.13 ° = 83.13 °$ and $30 ° + 126.87 ° = 156.87 °$ . Both sums are less than $180 °$ , so both candidates leave room for a third angle: $C = 96.87 °$ in the first triangle and $C = 23.13 °$ in the second.

\begin{array}{l} C = 180^{\circ} - 30^{\circ} - {53.13}^{\circ} \\ C = {96.87}^{\circ} \end{array}

Find side $c$ in each triangle

Use the law of sines once more with each value of $C$ . Because $5 / sin 30 ° = 10$ , each $c$ is $10 sin C$ : $9.93$ for $C = 96.87 °$ and $3.93$ for $C = 23.13 °$ .

\begin{array}{l} \frac{c}{sin {96.87}^{\circ}} = \frac{5}{sin 30^{\circ}} \\ c = \frac{5 sin {96.87}^{\circ}}{sin 30^{\circ}} \\ c \approx 9.93 \end{array}

The two triangles with $A = 30 °$ , $a = 5$ and $b = 8$
Part	First triangle	Second triangle
$B$	$53.13 °$	$126.87 °$
$C$	$96.87 °$	$23.13 °$
$c$	$9.93$	$3.93$

See why there are two

The height from $C$ is $b sin A = 4$ . Side $a = 5$ is longer than that height, so it reaches the opposite side, and shorter than $b = 8$ , so it can swing to either side of the height. The two triangles are those two positions.

Result

Two triangles: $B \approx 53.1 °$ , $C \approx 96.9 °$ and $c \approx 9.93$ , or $B \approx 126.9 °$ , $C \approx 23.1 °$ and $c \approx 3.93$ .

Your turn

1. One like the example

In triangle ABC, $A = 30 °$ , $b = 8$ and $a = 10$ . How many triangles fit, and what is $B$ ?

Show the answer and explanation

One triangle, with $B \approx 23.6 °$ .

$sin B$ $= 8 sin 30 ° / 10$ $= 0.4$ , so the candidates are $B \approx 23.58 °$ and $180 ° - 23.58 ° = 156.42 °$ . The second fails the test, because $30 ° + 156.42 °$ is more than $180 °$ . Only $B \approx 23.6 °$ gives a triangle.

\begin{array}{l} \frac{sin B}{8} = \frac{sin 30^{\circ}}{10} \\ sin B = \frac{8 sin 30^{\circ}}{10} \\ sin B = 0.4 \end{array}

2. A variation

In triangle ABC, $A = 40 °$ , $b = 8$ and $a = 3$ . How many triangles fit?

Show the answer and explanation

None. $sin B$ would be about $1.714$ , and no angle has a sine above $1$ .

$sin B$ $= 8 sin 40 ° / 3$ $\approx 8 (0.6428) / 3$ $\approx 1.714$ . Every sine is at most $1$ , so side $a$ is too short to reach the third side. The height from $C$ is $8 sin 40 ° \approx 5.14$ , longer than $a = 3$ .

\begin{array}{l} \frac{sin B}{8} = \frac{sin 40^{\circ}}{3} \\ sin B = \frac{8 sin 40^{\circ}}{3} \\ sin B \approx 1.714 \end{array}

3. Somewhere new

A triangular garden has a $110 °$ corner at $A$ . The side opposite that corner is $a = 9 m$ , and side $b$ is $6 m$ . How many triangles fit, and how long is side $c$ ?

Show the answer and explanation

One triangle, with $B \approx 38.8 °$ , $C \approx 31.2 °$ and $c \approx 4.96 m$ .

$sin B$ $= 6 sin 110 ° / 9$ $\approx 0.6265$ , so $B \approx 38.79 °$ or $141.21 °$ . The supplement fails, since $110 ° + 141.21 °$ is more than $180 °$ . That fits the shortcut, as $A$ is obtuse and $a$ is longer than $b$ . Then $C = 180 ° - 110 ° - 38.79 °$ $= 31.21 °$ , and $c = 9 sin 31.21 ° / sin 110 °$ $\approx 4.96 m$ .

\begin{array}{l} \frac{sin B}{6} = \frac{sin 110^{\circ}}{9} \\ sin B = \frac{6 sin 110^{\circ}}{9} \\ sin B \approx 0.6265 \end{array}

Keep exploring

Open $sin B$ in Math and change side $a$ from $5$ to $3$ in the first two rows, and the last row to $sin B = 4 / 3$ . Every row still reads ✓, because the arithmetic is right. Seeing that no angle has a sine above $1$ , so no triangle exists, is left to you.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 5 key terms Check sin B in Math Check side c of the first triangle in Math Check side c of the second triangle in Math Open worked example on a board Law of cosines in Math Reference

Your existing work stays on this device. Examples open as editable copies.

Recognize SSA

Find sinBsinB

List both candidates

Test each candidate

Find side cc in each triangle

See why there are two

Result

Your turn

1. One like the example

One triangle, with B≈23.6⁢°B≈23.6°.

2. A variation

None. sinBsinB would be about 1.7141.714, and no angle has a sine above 11.

3. Somewhere new

One triangle, with B≈38.8⁢°B≈38.8°, C≈31.2⁢°C≈31.2° and c≈4.96 mc≈4.96 m.

Keep exploring

Study it, then try it

Find $sin B$

Find side $c$ in each triangle

One triangle, with $B \approx 23.6 °$ .

None. $sin B$ would be about $1.714$ , and no angle has a sine above $1$ .

One triangle, with $B \approx 38.8 °$ , $C \approx 31.2 °$ and $c \approx 4.96 m$ .