If the roots of the equation $c^2x^2-c(a+b)x+ab=0$ are sin A, sin B where A, B and C are the angles and a, b, c are the opposite sides of a triangle, then the triangle is :

(i) Right angled

(ii) Acute angled

(iii) Obtuse angled

(iv) $\sin A+\cos A=\frac{a+b}{c}$

(i), (iv)

Sum of roots = $\sin A+\sin B=\frac{c(a+b)}{c^2}=\frac{\sin A+\sin B}{\sin C}$ [by using sine rule]

⇒ sin c = 1 ⇒ ∠C = 90° and A + B = 90° ⇒ cos A = sin B

Hence, $\sin A + \cos A = \sin A + \sin B =\frac{a+b}{c}$