Question

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

• A. is acute-angled
• B. is right-angled
• C. is obtuse- angled
• D. satisfies $\sin A+\cos A=\frac{a+b}{c}$

Answer 1

Answer: (B), (D)

The correct options are
B is right-angled
D satisfies $\sin A+\cos A=\frac{a+b}{c}$

$\because \sin A$ and $\sin B$ are the roots of
$c^2{x}^2-c(a+b)x+ab=0$
then $\sin A+\sin B=\frac{a+b}{c}$ (on simplification)
and $\sin A\sin B=\frac{ab}{c^2}$
Using, sine rule,
$\Rightarrow \frac{a}{2R}+\frac{b}{2R}=\frac{a+b}{c}$
$\Rightarrow \frac{a}{2R}\frac{b}{2R}=\frac{ab}{c^2}$
$\therefore c=2R$
$\Rightarrow 2R\sin C=2R$
$\Rightarrow \sin C=1$
$\therefore C=\frac{\pi }{2}$
Since $A+B+C=\pi$
$\Rightarrow A+B=\pi -C$
$\Rightarrow A+B=\pi -\frac{\pi }{2}=\frac{\pi }{2}$
and $\angle B=\frac{\pi }{2}-\angle A$
$\because \sin A+\sin B=\frac{a+b}{c}$
$\Rightarrow \sin A+\sin (\frac{\pi }{2}-\angle A)=\frac{a+b}{c}$
$\Rightarrow \sin A+\cos A=\frac{a+b}{c}$