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Question

If a2+b2+c2=8R2, then prove that the triangle is a right-angled triangle.


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Solution

Proof:

Given, a2+b2+c2=8R2

Using Sin law, asinA=bsinB=csinC=2R

(2RsinA)2+(2RsinB)2+(2RsinC)2=8R2

4R2sin2A+4R2sin2B+4R2sin2C=8R2

4R2sin2A+sin2B+sin2C=8R2

sin2A+sin2B+sin2C=2

1-cos2A2+1-cos2B2+1-cos2C2=2 cos2θ=1-2sin2θsin2θ=1-cos2θ2

3(cos2A+cos2B+cos2C)=4

cos2A+cos2B+cos2C=1

14cosAcosBcosC=1(Cos2A+Cos2B+Cos2C=-1-4cosAcosBcosC,whenA+B+C=180)

cosAcosBcosC=0

For the above equation to be 0, either of cosA or cosB or cosC should be equal to 0.

We know that, cos90°=0.

Thus, either of angles A,B or C is 90°(right-angled).

Hence, it is proved that the triangle is a right-angled triangle.


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