Math · Precalculus · Worked example

Solve an AAS triangle with the law of sines

In triangle ABC, $A = 45 °$ , $B = 60 °$ and $a = 6$ . Find angle $C$ and sides $b$ and $c$ .

Find the third angle

Two angles are known, so the angle sum gives the third.

\begin{array}{l} C = 180^{\circ} - 45^{\circ} - 60^{\circ} \\ C = 75^{\circ} \end{array}

Choose the law

Side $a$ and angle $A$ are a complete pair, so the law of sines applies. Pair each unknown side with its own opposite angle.

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}

Pair $b$ with its angle

Side $b$ faces the $60 °$ angle. Set $b$ over $sin 60 °$ equal to the complete pair, $6$ over $sin 45 °$ .

\frac{b}{sin 60^{\circ}} = \frac{6}{sin 45^{\circ}}

Solve for $b$

Multiply both sides by $sin 60 °$ , which leaves $b$ alone on the left.

b = \frac{6 sin 60^{\circ}}{sin 45^{\circ}}

Find the exact value

With $sin 60 ° = \sqrt{3} / 2$ and $sin 45 ° = \sqrt{2} / 2$ , the halves cancel and $b = 6 \sqrt{3} / \sqrt{2}$ . Multiplying the top and bottom by $\sqrt{2}$ gives $6 \sqrt{6} / 2 = 3 \sqrt{6}$ .

b = 3 \sqrt{6}

Round the length

$3 \sqrt{6} \approx 7.348$ , which rounds to $7.35$ .

b \approx 7.35

Find side $c$

Repeat with angle $C$ . A calculator gives $sin 75 ° \approx 0.9659$ .

\begin{array}{l} \frac{c}{sin 75^{\circ}} = \frac{6}{sin 45^{\circ}} \\ c = \frac{6 sin 75^{\circ}}{sin 45^{\circ}} \\ c \approx 8.20 \end{array}

Check the order

The longest side faces the largest angle: $c \approx 8.20$ faces $75 °$ , $b \approx 7.35$ faces $60 °$ and $a = 6$ faces $45 °$ . An answer that breaks this order has a side paired with the wrong angle.

Result

$C = 75 °$ , $b = 3 \sqrt{6} \approx 7.35$ and $c \approx 8.20$ .

Your turn

1. One like the example

In triangle ABC, $B = 30 °$ , $C = 45 °$ and $b = 4$ . Find side $c$ .

Show the answer and explanation

$c = 4 \sqrt{2} \approx 5.66$ .

Side $b$ and angle $B$ are a complete pair, so $c / sin 45 ° = 4 / sin 30 °$ . Then $c = 4 sin 45 ° / sin 30 °$ $= 4 (\sqrt{2} / 2) / (1 / 2)$ $= 4 \sqrt{2}$ .

\begin{array}{l} \frac{c}{sin 45^{\circ}} = \frac{4}{sin 30^{\circ}} \\ c = \frac{4 sin 45^{\circ}}{sin 30^{\circ}} \\ c = 4 \sqrt{2} \end{array}

2. A variation

In triangle ABC, $A = 40 °$ , $C = 75 °$ , and the side between those two angles is $b = 10$ . Find $B$ and side $a$ .

Show the answer and explanation

$B = 65 °$ and $a \approx 7.09$ .

This is ASA. The angle sum gives $B = 180 ° - 40 ° - 75 °$ $= 65 °$ , and now $b = 10$ and $B = 65 °$ are a complete pair. Then $a / sin 40 ° = 10 / sin 65 °$ , so $a = 10 sin 40 ° / sin 65 °$ $\approx 10 (0.6428) / 0.9063$ $\approx 7.09$ .

\begin{array}{l} \frac{a}{sin 40^{\circ}} = \frac{10}{sin 65^{\circ}} \\ a = \frac{10 sin 40^{\circ}}{sin 65^{\circ}} \\ a \approx 7.09 \end{array}

3. Somewhere new

Two lookouts stand $2 km$ apart at $A$ and $B$ on a straight shore. A boat at $C$ makes an angle of $52 °$ with the shore at $A$ and $61 °$ at $B$ . How far is the boat from $A$ ?

Show the answer and explanation

About $1.90 km$ .

The angle at the boat is $C = 180 ° - 52 ° - 61 °$ $= 67 °$ , and the shore $A B = 2 km$ faces it, a complete pair. The distance from $A$ to the boat is side $b$ , which faces the $61 °$ angle at $B$ . Then $b = 2 sin 61 ° / sin 67 °$ $\approx 2 (0.8746) / 0.9205$ $\approx 1.90 km$ .

\begin{array}{l} \frac{b}{sin 61^{\circ}} = \frac{2}{sin 67^{\circ}} \\ b = \frac{2 sin 61^{\circ}}{sin 67^{\circ}} \\ b \approx 1.90 \end{array}

Keep exploring

In the Math box for side $b$ , flip the ratio to $b = 6 sin 45 ° / sin 60 °$ , which pairs each side with the wrong angle. The row turns to ✗: that $b$ would be $2 \sqrt{6} \approx 4.90$ , shorter than a although it faces a larger angle.

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Find the third angle

Choose the law

Pair bb with its angle

Solve for bb

Find the exact value

Round the length

Find side cc

Check the order

Result

Your turn

1. One like the example

c=42≈5.66c=4√2≈5.66.

2. A variation

B=65⁢°B=65° and a≈7.09a≈7.09.

3. Somewhere new

About 1.90 km1.90 km.

Keep exploring

Study it, then try it

Pair $b$ with its angle

Solve for $b$

Find side $c$

$c = 4 \sqrt{2} \approx 5.66$ .

$B = 65 °$ and $a \approx 7.09$ .

About $1.90 km$ .