Law of Sines and Cosines

Beyond the Right Triangle

SOH-CAH-TOA only works when a triangle has a $90^\circ$ angle. But the ACT also asks about oblique triangles — triangles with no right angle — and for those you need two more tools: the law of sines and the law of cosines. Together they let you find any missing side or angle as long as you have enough information to pin the triangle down.

Throughout, we use the standard labeling: a lowercase letter for each side and the matching capital for the angle opposite it. So side $a$ is across from angle $A$, side $b$ from angle $B$, and side $c$ from angle $C$.

The Law of Sines

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

Each side divided by the sine of its opposite angle gives the same value. This is the tool of choice when you have a matched pair — a side and its opposite angle — plus one more piece. That covers two common setups:

• AAS / ASA — two angles and any side. Find the third angle from $A + B + C = 180^\circ$, then use the law of sines for the remaining sides.
• SSA — two sides and an angle opposite one of them.

Example. In a triangle $A = 40^\circ$, $B = 75^\circ$, and side $a = 10$. Since $a$ pairs with $A$, find $b$:

$\frac{10}{\sin 40^\circ} = \frac{b}{\sin 75^\circ} \;\Rightarrow\; b = \frac{10\sin 75^\circ}{\sin 40^\circ} \approx 15.0$

The Ambiguous Case (SSA)

When you know two sides and an angle not between them, there can be two different triangles that fit. After solving $\sin B = \dfrac{b\sin A}{a}$, remember that an obtuse angle with the same sine — namely $180^\circ - B$ — may also work. The ACT rarely demands both answers, but knowing the trap keeps you from assuming a unique triangle when the problem hints at two.

The Law of Cosines

$c^2 = a^2 + b^2 - 2ab\cos C$

This is the law of sines' partner for the cases sine can't start: when you have no side-angle pair to work from. Use it for:

• SAS — two sides and the included angle (the angle between them). Solve directly for the third side.
• SSS — all three sides. Rearrange to find any angle:

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Notice that when $C = 90^\circ$, $\cos C = 0$ and the formula collapses to $c^2 = a^2 + b^2$ — the Pythagorean theorem is just the law of cosines for a right angle.

Example (SAS). Two sides measure $7$ and $9$ with a $40^\circ$ angle between them. The opposite side is

$c^2 = 7^2 + 9^2 - 2(7)(9)\cos 40^\circ \approx 130 - 96.5 = 33.5,$

so $c \approx 5.8$.

Area from Two Sides and an Angle

A close cousin finishes many ACT geometry questions. When you know two sides and the included angle, the area is

$\text{Area} = \tfrac{1}{2}ab\sin C.$

For sides $10$ and $14$ with an included $45^\circ$ angle, the area is $\tfrac{1}{2}(10)(14)\sin 45^\circ \approx 49.5$.

Choosing the Right Law

What you know	Tool
A side and its opposite angle (AAS, ASA, SSA)	Law of sines
Two sides + the included angle (SAS)	Law of cosines
All three sides (SSS)	Law of cosines (rearranged for the angle)
Two sides + the included angle, need area	$\tfrac{1}{2}ab\sin C$

The quickest way to decide: do I already have a side matched with its opposite angle? If yes, sines; if no, cosines.

Common Mistakes to Avoid

• Reaching for the law of sines on an SAS or SSS triangle, where no side-angle pair exists yet.
• Forgetting that $\cos C$ is negative for an obtuse angle, which makes the $-2ab\cos C$ term add length.
• Pairing the wrong side with the wrong angle — the side must be opposite the angle you use.
• Overlooking the second possible triangle in an SSA (ambiguous) situation.