Question

In Triangle ABC, A=60°. Prove that BC^2=AB^2+ AC^2- AB· AC.

Answer 1

$\mathbb\red{Given:-}$

• A = 60° in ∆ABC.

$\mathbb\red{Find:-}$

•$\sf Prove \: that \: BC^2 \: =AB^2 \: + \: AC^2- \: AB· \: AC.$

$\mathbb\red{Solution:-}$

•Applying cosine rule to △ABC,

$\sf\rightarrow cos A = \frac{ {ab}^{2} + {ac}^{2} - {bc}^{2} }{2.ab.ac.ac}$

$\sf\rightarrow cos 60 = \frac{ {ab}^{2} + {ac}^{2} - {bc}^{2} }{2.ab.ac}$

$\sf\rightarrow \: \frac{1}{2} = \frac{ {ab}^{2} + {ac}^{2} - {bc}^{2} }{2.ab.ac}$

$\therefore BC^2 = AB^2 + AC^2 - AB.AC$