In a $\triangle A B C, \sin A$ and $\sin B$ satisfy $c^2 x^2-c(a+b) x+a b=0$, then

• A

  the triangle is acute angled
• B

  the triangle is obtuse angled
• C

  $\sin C=\frac{\sqrt{3}}{2}$
• Correct answer
  D

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

Solution

We have, $$ \begin{array}{rrrl} & & c^2 x^2-c(a+b) x+a b=0 \\ \Rightarrow & c^2 x^2-c a x-c b x+a b=0 \\ \Rightarrow & c x(c x-a)-b(c x-a)=0 \\ \Rightarrow & & (c x-a)(c x-b)=0 \\ \Rightarrow & & c x-a=0 \text { or } c x-b=0 \\ \Rightarrow & x=\frac{a}{c} \text { and } x=\frac{b}{c} \end{array} $$ As, $\sin A$ and $\sin B$ are roots of the given equation. Let $\sin A=\frac{a}{c}$ and $\sin B=\frac{b}{c}$ $\Rightarrow \quad c=\frac{a}{\sin A}$ and $c=\frac{b}{\sin B}$ $\Rightarrow \quad \frac{a}{\sin A}=\frac{b}{\sin B}=c$ Using sine law, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ So, $\quad \frac{c}{\sin c}=c \quad$ [from Eqs. (i) and (ii)] $$ \begin{aligned} & \Rightarrow \quad \sin c=\frac{c}{c} \\ & \Rightarrow \quad \sin C=1 \quad\left[\because \sin \frac{\pi}{2}=1\right] \\ & \Rightarrow \quad \angle C=\frac{\pi}{2} \\ & \end{aligned} $$ Since, $\triangle A B C$ is right angled at $\angle c$. $$ \text { In } \triangle A B C \text {, } $$
$$ \begin{aligned} \sin A & =\frac{a}{c} \text { and } \cos A=\frac{b}{c} \\ \therefore \quad \sin A+\cos A & =\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c} \end{aligned} $$

Hence, the correct answer is option: 'D'.