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