Q.In $\triangle ABC$ prove that $\dfrac{c - b\cos A}{b - c\cos A} = \dfrac{\cos B}{\cos C}$.

Cosine Rule

The sine rule needs at least one known angle paired with its opposite side to get started. But two very common situations don't offer that: you might know two sides and the angle between them (SAS), or all three sides (SSS) with no angle at all. The Cosine Rule is built exactly for these cases:

and cyclically for $B$ and $C$. Think of it as a generalisation of the Pythagorean theorem: when $A=90^\circ$, $\cos A=0$ and the formula collapses to $a^2=b^2+c^2$ exactly.

Where it comes from

Placing $A$ at the origin with $B$ along the $x$-axis, the third vertex $C$ sits at $(b\cos A,b\sin A)$ purely from the definition of angle $A$ and the length $AC=b$. The distance formula between $B=(c,0)$ and $C$ then expands, via $\cos^2A+\sin^2A=1$, straight into $a^2=b^2+c^2-2bc\cos A$ — no trigonometric identity beyond the Pythagorean one is needed.

How it is used

• SAS → third side: given $b,c,A$, compute $a$ directly, then use the sine rule to find $B$ or $C$ (picking the smaller of the two unknown angles avoids the ambiguous-angle trap).
• SSS → any angle: solve the rearranged form for $\cos A$ (or $B$, or $C$) using all three known sides, then take $\cos^{-1}$.
• A building block: the half-angle formulas, and Heron's formula, are both derived from the cosine rule by combining it with the double-angle identities $\cos A=1-2\sin^2\frac A2=2\cos^2\frac A2-1$. …