Trigonometry Reference
Law of Sines and Cosines
These formulas solve non-right triangles. The method depends on what information is given: matching side-angle pairs point to the Law of Sines, while SAS and SSS point to the Law of Cosines.
Fact Table
| Given | Usually Use | What to Do First |
|---|---|---|
| AAS | Law of Sines. | Find the third angle, then use the known side-angle pair. |
| ASA | Law of Sines. | Find the third angle if needed, then set up a proportion. |
| SAS | Law of Cosines. | Find the side across from the included angle. |
| SSS | Law of Cosines. | Find an angle from the three side lengths. |
| SSA | Law of Sines. | Check the supplement because there may be zero, one, or two triangles. |
Content Formulas
Law of Sines
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$
Law of Cosines: Side
$$c^2=a^2+b^2-2ab\cos C$$
Law of Cosines: Angle
$$C=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)$$
Angle Sum
$$A+B+C=180^\circ$$
Lowercase sides sit across from matching uppercase angles: side $a$ is opposite angle $A$, side $b$ is opposite angle $B$, and side $c$ is opposite angle $C$.
Classic Examples
AAS: Find a Side
In triangle $ABC$, $A=40^\circ$, $B=70^\circ$, and $a=20$. Find $b$.
Solution MoveUse the Law of Sines because one complete pair, $A$ and $a$, is known.
$$\frac{20}{\sin 40^\circ}=\frac{b}{\sin 70^\circ}$$
$$b=\frac{20\sin 70^\circ}{\sin 40^\circ}$$
$$b\approx 29.2$$
ASA: Find the Third Angle
In triangle $ABC$, $A=55^\circ$, $C=75^\circ$, and $b=12$. Find $a$.
Solution MoveFind the missing angle first, then use the known pair $B$ and $b$.
$$B=180^\circ-55^\circ-75^\circ$$
$$B=50^\circ$$
$$\frac{a}{\sin 55^\circ}=\frac{12}{\sin 50^\circ}$$
$$a=\frac{12\sin 55^\circ}{\sin 50^\circ}$$
$$a\approx 12.8$$
SAS: Find a Side
Two sides of a triangle are 10 and 15, and the included angle is $60^\circ$. Find the opposite side.
Solution MoveUse the Law of Cosines because the known angle is between the two known sides.
$$c^2=10^2+15^2-2(10)(15)\cos 60^\circ$$
$$c^2=100+225-150$$
$$c^2=175$$
$$c\approx 13.2$$
SSS: Find an Angle
A triangle has side lengths 10, 15, and 20. Find the angle across from the side of length 20.
Solution MoveUse the angle form of the Law of Cosines. Put the target side opposite the target angle.
$$C=\cos^{-1}\left(\frac{10^2+15^2-20^2}{2(10)(15)}\right)$$
$$C=\cos^{-1}\left(\frac{-75}{300}\right)$$
$$C=\cos^{-1}(-0.25)$$
$$C\approx 104.5^\circ$$
SSA: Two Triangles
In triangle $ABC$, $A=30^\circ$, $a=10$, and $b=15$. Find the possible values of $B$.
Solution MoveAfter inverse sine, test the supplement. Keep it if the angle sum stays below $180^\circ$.
$$\frac{10}{\sin 30^\circ}=\frac{15}{\sin B}$$
$$10\sin B=15\sin 30^\circ$$
$$\sin B=0.75$$
$$B\approx 48.6^\circ$$
The sine value is also positive at the supplement.
$$180^\circ-48.6^\circ=131.4^\circ$$
$$30^\circ+131.4^\circ=161.4^\circ$$
$$B\approx 48.6^\circ\quad \text{or}\quad B\approx 131.4^\circ$$