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

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

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

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

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

Two sides of a triangle are given by the roots of the equation x^(2)-5x+6=0 and the angle between the sides is 60 ° Then the perimeter of the triangle is a

Two sides of a triangle are given by the roots of the equation x^(2)-5x+6=0 and the angle between the sides is (pi)/(3). Then the perimeter of the triangle is

Two sides of a triangle are given by the roots of the equation x^(2) - 5x+6 =0 and the angle between the sides is (pi)/(3). Then the perimeter of the triangle is-