If two sides of a triangle are roots of the equation `x^2-7x+8=0` and the angle between these sides is `60^0` then the product of inradius and circumradius of the triangle is `8/7` (b) `5/3` (c) `(5sqrt(2))/3` (d) `8`

Explanation

`x^2-7x+8=0`

a+b=7

ab=8

`/_=60^o`

`cos60^@=(a^2+b^2-c^2)/(2ab)`

`1/2=((a+b)^2-2ab-c^2)/(2ab)`

`1/2=(49-2*8-c^2)/(2*8)`

`c^2=25`

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