Question

In $△$ ABC, $\angle A=60$. Prove that ${BC}^2={AB}^2+{AC}^2-AB\cdot AC$.

Answer 1

Given $\angle A={60}^{\circ }$ in $△ABC$.

Applying cosine rule to $△ABC$,

$\cos A=\frac{A{B}^2+A{C}^2-B{C}^2}{2.AB.AC}$

$\cos 60=\frac{A{B}^2+A{C}^2-B{C}^2}{2.AB.AC}$

$\frac{1}{2}=\frac{A{B}^2+A{C}^2-B{C}^2}{2.AB.AC}$

$\therefore B{C}^2=A{B}^2+A{C}^2-AB.AC$