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Q.

If twice the square of the radius of a circum circle is equal to half the sum of the squares of the sides of inscribed triangle ABC, then  cos2A+cos2B+cos2C=

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a

2

b

8

c

1

d

4

answer is B.

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Detailed Solution

Give   2R2=12(a2+b2+c2)4R2=4R2[sin2A+sin2B+sin2C]sin2A+sin2B+sin2C=11−cos2A+1−cos2B+1−cos2C=13−1=cos2A+cos2B+cos2C∴cos2A+cos2B+cos2C=2

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