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.

Solved Examples

Jump directly to a worked example by triangle information type.

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 Steps
  1. Use the Law of Sines because the complete pair $A$ and $a$ is known.
  2. Solve the proportion for $b$.
  3. State the approximate side length: $b\approx29.2$.
$$\frac{20}{\sin 40^\circ}=\frac{b}{\sin 70^\circ}$$ $$b=\frac{20\sin 70^\circ}{\sin 40^\circ}$$ $$b\approx 29.2$$

ASA: Find the Third Angle

In triangle $ABC$, $A=55^\circ$, $C=75^\circ$, and $b=12$. Find $a$.

Solution Steps
  1. Use the angle sum to find the missing angle.
  2. State $B=50^\circ$.
  3. Set up the Law of Sines using the known pair $B$ and $b$.
  4. Solve the proportion for $a$.
  5. State the approximate side length: $a\approx12.8$.
$$B=180^\circ-55^\circ-75^\circ$$ $$B=50^\circ$$ $$\frac{a}{\sin 55^\circ}=\frac{12}{\sin 50^\circ}$$ $$a=\frac{12\sin 55^\circ}{\sin 50^\circ}$$ $$a\approx 12.8$$

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 Steps
  1. Use the Law of Cosines because the known angle is included between the known sides.
  2. Evaluate the cosine expression.
  3. Simplify the square of the target side.
  4. Take the positive square root and state $c\approx13.2$.
$$c^2=10^2+15^2-2(10)(15)\cos 60^\circ$$ $$c^2=100+225-150$$ $$c^2=175$$ $$c\approx 13.2$$

SSS: Find an Angle

A triangle has side lengths 10, 15, and 20. Find the angle across from the side of length 20.

Solution Steps
  1. Use the angle form of the Law of Cosines with the target side opposite $C$.
  2. Evaluate the numerator and denominator of the fraction.
  3. Evaluate the inverse cosine input.
  4. State the approximate angle: $C\approx104.5^\circ$.
$$C=\cos^{-1}\left(\frac{10^2+15^2-20^2}{2(10)(15)}\right)$$ $$C=\cos^{-1}\left(\frac{-75}{300}\right)$$ $$C=\cos^{-1}(-0.25)$$ $$C\approx 104.5^\circ$$

SSA: Two Triangles

In triangle $ABC$, $A=30^\circ$, $a=10$, and $b=15$. Find the possible values of $B$.

Solution Steps
  1. Set up the Law of Sines with the known side-angle pair.
  2. Isolate $\sin B$.
  3. Evaluate the sine value.
  4. Find the first possible angle with inverse sine.
  5. Find its supplement.
  6. Check that the angle sum remains below $180^\circ$.
  7. State both possible values: $B\approx48.6^\circ$ or $B\approx131.4^\circ$.
$$\frac{10}{\sin 30^\circ}=\frac{15}{\sin B}$$ $$10\sin B=15\sin 30^\circ$$ $$\sin B=0.75$$ $$B\approx 48.6^\circ$$ $$180^\circ-48.6^\circ=131.4^\circ$$ $$30^\circ+131.4^\circ=161.4^\circ$$ $$B\approx 48.6^\circ\quad \text{or}\quad B\approx 131.4^\circ$$