If in a - ABC - a 2- b 2- c 2-8 R 2- where R - circumradius- then the triangle isA- equilateralB- isoscelesC- right angledD- No such triangle exists

The correct option is C right angled a^2+b^2+c^2=8R^2 ⇒ sin^2A+sin^2B+sin^2C=2 ⇒ cos2A+cos2B+cos2C+1=0 ∴ 4cosA.cosB.cosC=0 ⇒ A or B or C=π/2 ...

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Value of Standard Angle(30,60,90)

If in a Δ ABC...

Question

If in a $Δ$ ABC, $a^{2} + b^{2} + c^{2} = 8 R^{2},$ where R = circumradius, then the triangle is

equilateral

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isosceles

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right angled

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No such triangle exists

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