Application to Triangle

Solving a triangle means computing every one of its six elements -- the three sides and the three angles -- once enough of them are already known. A right triangle needs only two elements besides the right angle itself (at least one of the two being a side): the third side follows from the Pythagorean theorem, and the missing acute angle follows because the two acute angles of a right triangle always add up to $90^\circ$. An oblique (non-right) triangle instead needs three known elements, of which at least one must be a side -- knowing only angles fixes a triangle's shape but never its size, since infinitely many similar triangles share the same three angles.

Working rule -- which formula to reach for

• All three sides known (SSS). Use the cosine rule (or the half-angle formulas) to compute every angle -- read one angle off $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ and its cyclic versions for $B$ and $C$ -- then use the angle sum $A+B+C = 180^\circ$ as a check.
• Two angles and a side opposite one of them known (SAA/ASA). Find the third angle from the angle sum, then the sine rule $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ gives the two remaining sides directly.
• Two sides and the included angle known (SAS). The sine rule cannot be used yet, because no side's opposite angle is known; instead the cosine rule finds the third side first. Once all three sides are known the problem reduces to the SSS case above for the remaining angles. A SAS triangle is always unique.
• Every one of these methods needs at least one side length pinned down -- that is exactly what turns "the shape of a triangle" into "this one particular triangle".

The five-case classification

* SSA means two sides and a non-included angle.

The SSA ambiguous case, worked out. Suppose $a$, $b$ and $A$ are known, with $A$ not the angle between the two known sides. Let $h = b\sin A$ be the perpendicular height dropped from the vertex opposite side $b$ onto the base line, and think of side $a$ as a swinging arm of fixed length pivoting to try to close the triangle. Comparing that swinging length $a$ against $h$ and against $b$ decides everything:

• If $a < h$: the arm is too short to even reach the base line -- no triangle.
• If $a = h$: the arm just touches the base line at a right angle -- exactly one (right) triangle.
• If $h < a < b$: the arm crosses the base line twice on the same side of the starting vertex -- two triangles (the genuinely ambiguous case). …