Question

If $\sin A$ and $\sin B$ of a triangle $ABC$ satisfy $c^2{x}^2-c(a+b)x+ab=0$, then the triangle is

• A. Equilateral
• B. Isosceles
• C. Right angled
• D. Acute angled

Answer 1

The correct option is C Right angled

$\therefore \sin A+\sin B=\frac{C(a+b)}{C^2}=\frac{a+b}{c}$

$\therefore \sin A\sin B=\frac{ab}{c^2}$

Also By rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$\therefore \sin A\sin B=\frac{ab}{k^2}=\frac{ab}{c^2}\to$ Given

$C^2{x}^2-C(a+b)x+ab=0(\sin A,\sin B)$

$\therefore k=c$

$\therefore \frac{a+b}{c}=\frac{k(\sin A+\sin B)}{k\left(\sin c\right)}=\frac{\sin A+\sin B}{\sin C}$

$k=2R=C$

Right angled triangle