Correct Answer – C
We have,
`r_(1) = (Delta)/(s-a) = 2, r_(2) = (Delta)/(s-b) = 3, r_(3) = (Delta)/(s-c) = 6`
Given `Delta =6`,
`:. S-a = 3` …(i)
`s-b =2` …(ii)
`s-c =1` …(iii)
Adding Eqs. (i) and (ii), `2s – a – b =5 ” or ” a + b + c -a -b = 5 or c = 5`
Adding Eqs. (i) and (iii), `2s – a- c = 4, or b = 4`
And adding Eqs. (ii) and (iii), `2s -b -c = 3 or a = 3`
Hence the sides of the `Delta` are `a =3, b = 4, c = 5`
Since the triangle is right angled, the greatest angle is `90^(@)`.
Also, the least angle is opposite to side a, which is `sin^(-1).(3)/(5)`. Therefore,
`90^(@) – “sin”^(-1)(3)/(5) = “cos”^(-1)(3)/(5)`
Also, `R = (abc)/(4Delta) = (60)/(24) = 2.5`
`r = (Delta)/(s) = (6)/(6) = 1`