Question

In ∆ABC, ∠A = 60°. Prove that BC² = AB² + AC² − AB . AC.

Answer 1

In ΔABC, in which ∠A is an acute angle with 60°.

`sin 60^o = (CD)/(AC)=sqrt3/2`

`⇒ CD = sqrt3/2AC`.................(1)

`cose 60^o = (AD)/(AC)=1/2`

`⇒ AD = 1/2 AC`

Now apply Pythagoras' theorem in triangle BCD

`BC^2=CD^2+BD^2`

`= CD^2 +(AB-AD)^2`

`= (sqrt3/2AC)^2+AB^2+(1/2AC)^2-2AB1/2AC`

`=AC^2+AB^2-AB.AC`

Hence `BC^2=AB^2+AC^2-AB.AC`