The sides of a triangle are `sin alpha, cos alpha, sqrt(1+sin alpha cos alpha)" for some " 0 lt alpha lt (pi)/(2)` then the greatest angle of the triangle is :

Options

A. `150^(@)`
B. `90^(@)`
C. `120^(@)`
D. `60^(@)`

Explanation

Let ABC be a triangle such that
a= sin `alpha`, b = cos `alpha` and `c=sqrt(1+sinalphacosalpha)` where `0ltalphalt(pi)/2`
Clearly, 0 lt a lt 1, 0 lt b lt 1 and c gt 1. Therefore, c is the greatest side and hence angle C is the greatest angle.
Now,
`cosC=(a^(2)+b^(2)-c^(2))/(2ab)`
`rArrcosC=(sin^(2)alpha+cos^(2)alpha-(1+sinalphacosalpha))/(2sinalphacosalpha)`
`rArrcosC=-1/2rArrC=120^(@)`