If the angles `alpha, beta, gamma` of a triangle satisfy the relation,
`sin((alpha-beta)/(2)) + sin ((alpha -gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)`, then
The measure of the smallest angle of the triangle is

Options

A. `30^(@)`
B. `40^(@)`
C. `45^(@)`
D. `50^(@)`

Answer

B

Explanation

We have `sin((alpha - beta)/(2)) + sin ((alpha - gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)`
`therefore sin (((pi-(beta+gamma))-beta)/(2))+ sin (((pi-(gamma +beta))-gamma)/(2)) + sin((3pi-3(beta+gamma))/(2)) = (3)/(2)`
`therefore cos((2beta + gamma)/(2)) + cos((2gamma + beta)/(2)) - cos((3(beta+gamma))/(2)) = (3)/(2)`
`therefore 2cos ((3)/(4)(beta + gamma)) cos ((beta-gamma)/(4)) +1 - 2cos^(2)((3)/(4)(beta +gamma)) = (3)/(2)`
`therefore 4 cos^(2) ((3)/(4) (beta + gamma)) - 4cos((3)/(4) (beta + gamma)) cos ((beta-gamma)/(4)) +1 =0`...(1)
Above equation is quadratic in `cos((3)/(4)(beta +gamma))`
Since `cos((3)/(4)(beta+gamma))` is a real number,
Discriminant `D ge 0`
`therefore 16 cos^(2)((beta-gamma)/(4)) -16 ge 0`
`rArr cos ^(2)((beta -gamma)/(4)) ge 1`
`rArr cos^(2)((beta -gamma)/(4)) =1 `
`rArr beta = gamma `
From equation (1) for `beta = gamma`, we get
`[ 2 cos((3)/(4)(beta+gamma)) -1]^(2) =0`
`rArr 2cos""(3)/(4) (beta + gamma) = 1`
`rArr cos""(3)/(4) (beta +gamma) = (1)/(2)`
`rArr (3)/(4) (beta +gamma) = 60^(@)`
`therefore beta +gamma = 80^(@)`
`therefore alpha = 100^(@)`
`therefore beta = 40^(@), gamma = 40^(@)`