If the sines of the angles A and B of a triangle ABC satisfy the equation `c^2x^2-c(a+b)x+a b=0` , then the triangle a)acute angled b)right angled c)obtuse angled d)`sinA+cosA`= `((a+b))/c`

Options

A. is acute angled
B. is right angled
C. is obtus angled
D. satisfies the equation `sin A + cos A = ((a +b))/(c)`

Answer

B, D

Explanation

Since sin A and sin B are the roots of `c^(2) x^(2) - c (a+b) x + ab = 0`, we have
`sin A + sin B = (a+b)/(c) and sin A sin B = (ab)/(c^(2))`
`rArr (a)/(2R) + (b)/(2R) = (a+b)/(c) and (a)/(2R) xx (b)/(2R) = (ab)/(c^(2))`
`:. c = 2R`
`rArr 2R sin C = 2R`
`rArr angleC = 90^(@)`
`rArr A + B = 90^(@)`
`rArr B = 90^(@) - A`
`:' sin A + sin B = (a+b)/(c)`
`:. x^(2) + (1)/(x^(2)) ge 2`