1.

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`

Answer» `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`
`c=5`
`r*R=D/S*(abc)/(4D)=(8*5)/(4(a+b+c)/2)`
`r*R=5/3`.


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