The Sine Law

Take any triangle $ABC$ and drop a perpendicular (the altitude $h$) from $C$ down to side $c$. It splits the triangle into two right triangles, so SOH CAH TOA applies to each. In the left one, $h$ is opposite angle $A$ and $b$ is the hypotenuse, so $\sin A=\frac{h}{b}$, giving $h=b\sin A$. In the right one, $h$ is opposite angle $B$ and $a$ is the hypotenuse, so $h=a\sin B$. Both expressions equal the same height, so they equal each other. Dividing both sides by $\sin A\sin B$ gives the first half of the sine law. Dropping the altitude from a different vertex brings in $c$ and $\sin C$ the same way. If the triangle is obtuse, the altitude falls outside, but because $\sin (180°-B)=\sin B$ the same equation still comes out.

Proof: Law of sines | Trig identities and examples | Trigonometry | Khan Academy

The sine law needs at least one complete angle/opposite-side PAIR that you already know. It applies to two classic situations: ASA or AAS (two angles and a side — find the third angle by subtracting from $180°$, then use the sine law for the remaining sides), and SSA (two sides and a non-included angle — the trickier 'ambiguous case,' covered in the next topic). When you are solving for an ANGLE, flip every fraction first so the unknown is on top: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$. This form is equally true and saves an algebra step.

1. Sketch and label the triangle: lower-case letters for sides, matching upper-case letters for the OPPOSITE angles.
2. If two angles are known, find the third by subtracting from $180°$ (angles in a triangle always sum to $180°$).
3. Use only two of the three ratios at a time (a matched pair you know, and the unknown pair) — cross-multiply to solve.
4. For a side, put sides on top: $b=\frac{a\sin B}{\sin A}$. For an angle, put sines on top: $\sin B=\frac{b\sin A}{a}$, then apply ${\sin }^{-1}$.
5. Check: the longest side must be opposite the largest angle, and the shortest side opposite the smallest angle.

One trap deserves its own warning. For the positive ratios that come up in a triangle, the inverse sine key only returns an angle between $0°$ and $90°$, yet a triangle can contain an obtuse angle, and $\sin B$ has the same value for $B$ and $180°-B$. So when the sine law hands you an ANGLE, ask whether the obtuse partner could also fit. Two quick tests settle it: if the side opposite the unknown angle is shorter than the known side in your pair, the unknown angle must be smaller than the known angle, so it cannot be obtuse; and if adding the obtuse partner to the known angle already reaches $180°$, it is impossible. A safe habit when you have a choice is to use the sine law for the SMALLEST unknown angle (opposite the shortest side), since that angle is always acute, then finish with the angle sum. This is exactly how forest-fire lookouts triangulate smoke: two towers a known distance apart each measure an angle, the third angle comes from the angle sum, and the sine law gives the distance from each tower to the fire.

Sine Law Explained: How It Works + 2 Examples

A 3D problem is really two (or more) flat triangles standing in different planes that share one side. The classic setup: two observers on level ground look at the base of a tower (a horizontal, oblique triangle) and one of them measures the angle of elevation to the top (a vertical, right triangle). Redraw each triangle flat on its own, mark which angles are horizontal and which are vertical, and solve the triangle that already has enough information first. The side it produces, often the ground distance from an observer to the base, is then carried into the other triangle. The sine law usually handles the ground triangle; SOH CAH TOA finishes the vertical one. A labelled sketch of each triangle is worth more here than anywhere else in the unit, because a 3D diagram on paper distorts angles: a right angle at the base of a tower rarely LOOKS like $90°$ in a perspective drawing.