In an acute angled triangle `A B C ,r+r_1=r_2+r_3a n d/_B >pi/3,` then `b+2c<2a<2b+2c` `b+4c<4a<2b+4c` `b+4c<4a<4b+4c` `b+3c<3a<3b+3c`

Options

A. `b + 2c lt 2a lt 2b + 2c`
B. `b + 4cc lt 4a lt 2b + 4c`
C. `b + 4c lt 4a lt 4b + 4c`
D. `b + 3c lt 3a lt 3b + 3c`

Answer

D

Explanation

`r - r_(2) = r_(3) -r_(1)`
`rArr (Delta)/(s) - (Delta)/(s-b) = (Delta)/(s-c) -(Delta)/(s-a)`
or `(-b)/(s(-b)) = (c-a)/((s-a) (s-c))`
or `((s-a) (s-c))/(s(s-b)) = (a-c)/(b)`
`rArr tan^(2).(B)/(2) = (a-c)/(b)`
But `(B)/(2) in ((pi)/(6), (pi)/(4))`. Therefore,
`tan^(2).(B)/(2) in ((1)/(3),1)`
`rArr (1)/(3) lt (a-c)/(b) lt 1`
or `b lt 3a -3c lt 3b`
or `b + 3c lt 3a lt 3b + 3c`