Q.In $\triangle ABC$ if $\dfrac{\cos A}{a} = \dfrac{\cos B}{b}$ then show that it is an isosceles triangle.

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 $\cos A=1-2\sin^2\frac A2=2\cos^2\frac A2-1$. …