Question

In $\Delta A B C$, the value of $\angle A$ is obtained from the equation $3 \cos A+2=0 .$
The quadratic equation, whose roots are $\sin A$ and $\tan A$, is

• A. $3 x^{2}+\sqrt{5} x-5=0$
• B. $6 x^{2}-\sqrt{5} x-5=0$
• C. $6 x^{2}+\sqrt{5} x-5=0$
• D. $6 x^{2}+\sqrt{5} x+5=0$

Answer 1

Answer: (C)

We have,
$3 \cos A+2=0$
$\Rightarrow \cos A=\frac{-2}{3}$
$\therefore \sin A=\sqrt{1-\cos ^{2} A}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$
and $\tan A=\frac{\sin A}{\cos A}=\frac{\sqrt{5} / 3}{-2 / 3}=\frac{-\sqrt{5}}{2}$
Since, $\sin A$ and $\tan A$ are the roots of required equation.
Hence, equation can be written as
$x^{2}-(\sin \,A+\,\tan \,A) x+\sin\,A \tan \,A=0$
$\Rightarrow x^{2}-\left(\frac{\sqrt{5}}{3}-\frac{\sqrt{5}}{2}\right) x+\left(\frac{\sqrt{5}}{3}\right)\left(\frac{-\sqrt{5}}{2}\right)=0$
$\Rightarrow x^{2}+\frac{\sqrt{5}}{6} x-\frac{5}{6}=0$
$\Rightarrow 6 \,x^{2}+\sqrt{5} \,x-5=0$