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`
Correct Answer – B
Let a and b the roots of `x^(2) -7 x + 8 = 0`
Then `a + b = 7, ab = 8`
Also, `C = 60^(@)`
`rArr c^(2) = a^(2) + b^(2) – ab`
`rArr c^(2) = (a +b)^(2) -3ab = 49 -24 = 25`
`rArr c = 5`
`:. r.R = (abc)/(2(a+b+c)) = (8xx5)/(2(7+5)) = (5)/(3)`
`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`.