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If a=sqrt(3),b=1/2(sqrt(6)+sqrt(2))dot a...

If `a=sqrt(3),b=1/2(sqrt(6)+sqrt(2))dot` and `c=2,` then find `/_A`

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Verified by Experts

Using Cosine Rule
`cos A = (b^(2) + c^(2) -a^(2))/(2bc)`
`= ((1//4) (8 + 4 sqrt3) + 2 - 3)/(sqrt12 + sqrt4) = (1 + sqrt3)/(2(1 + sqrt3)) = (1)/(2)`
So, `A = (pi)/(3)`
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CENGAGE PUBLICATION-PROPERTIES AND SOLUTIONS OF TRIANGLE-Illustration
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