In ΔABC,if 8R2=a2+b2+c2,then the triangle is

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In ΔABC,if 8R2=a2+b2+c2,then the triangle is

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Text solutionVerified

8R2=a2+b2+c2=4R2(sin2A+sin2B+sin2C) ⇒ sin2A+sin2B+sin2C=2 ⇒ (cos2A−sin2C)+cos2B=0 ⇒ cos(A−C)cos(A+C)+cos2B=0 Þ 2cosAcosBcosC=0 So that, cosA=0 or cosB=0 or cosC=0 ⇒ A=π2 or B=π2 or C=π2.

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Question Text	In ΔABC,if 8R2=a2+b2+c2,then the triangle is
Answer Type	Text solution:1
Upvotes	150

In ΔABC,if 8R2=a2+b2+c2,then the triangle is

Text solutionVerified

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