In any `!ABC `, sin `A/2` is

Options

A. less than `(b+c)/a`
B. less than or equal to `a/(b+c)`
C. greater than `(2a)/(a+b+c)`
D. none of these

Explanation

We have,
`a/(b+c)=(sinA)/(sinB+sinC)=(2sinA/2cosA/2)/(2sin(B+C)/2cos(B-C)/2)=(sinA/2)/(cos(B-C)/2)`
`thereforesinA/2=a/(b+c)cos(B-C)/2lea/(b+c)`
So, option (b) is correct.
Now,
`(2a)/(a+b+c)=(2sinA)/(sinA+sinB+sinC)`
`rArr(2a)/(a+b+c)=(4sinA/2cosA/2)/(4cosA/2cosB/2cosC/2)=(sinA/2)/(cosB/2cosC/2)`
`rArrsinA/2=(2a)/(a+b+c)cosB/2cosC/2le(2a)/(a+b+c)`