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).
• If $a \ge b$: the arm crosses the base line only once -- exactly one triangle.

The same swinging-arm logic applies whichever letters carry the SSA data (e.g. a known angle $B$ together with sides $a$ and $b$ instead of $A$, $a$, $b$) -- just relabel which side plays the role of the "swinging" side (opposite the known angle) and which plays the role of the "adjacent" side used to build $h$, before comparing.

What the chapter's worked examples (Examples 3.64-3.71) illustrate

• SSS gives all three cosines. With all three sides given, the cosine-rule formula for $\cos A$, $\cos B$, $\cos C$ is applied three times (cyclically permuting the sides) -- no sine rule is needed at all.
• SAA gives the other two sides. Two known angles fix the third via the angle sum; the sine rule then hands over both remaining sides in one shot, since every ratio $\dfrac{\text{side}}{\sin(\text{opposite angle})}$ is the same constant.
• SAS gives the third side, then the angles. The included angle between two known sides lets the cosine rule produce the third side; the remaining angles then follow from the cosine rule again (care is needed if using the sine rule instead, since it can return an acute angle when the true angle is obtuse).
• Heron's formula reads the area straight off SSS. Once the semi-perimeter $s = \tfrac12(a+b+c)$ is known, the area follows without ever computing an angle: $\triangle = \sqrt{s(s-a)(s-b)(s-c)}$.
• An area identity. Combining the projection-style identity $a\cos A + b\cos B + c\cos C = 2a\sin B\sin C$ with the area formula $\triangle = \tfrac12 ac\sin B = \tfrac12 ab \sin C$ (solved for $\sin B$ and $\sin C$ in terms of $\triangle$) collapses the left-hand side down to the compact form $\dfrac{8\triangle^2}{abc}$.
• Triangulating a position from two known distances (the cell-tower style problem). Two known distances from a signal source to two fixed points a known distance apart form an SSS triangle; the cosine rule recovers the bearing angle at the source, and a follow-up right-triangle sine step converts that angle into a perpendicular offset. …