Question

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

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
$x^2-(\sin A+\tan A)x+\sin A\tan A=0$
$\Rightarrow x^2-\left(\frac{\sqrt{5}}{3}-\frac{\sqrt{5}}{2}\right)x+\left(\frac{\sqrt{5}}{3}\right)\left(\frac{-\sqrt{5}}{2}\right)=0$
$\Rightarrow x^2+\frac{\sqrt{5}}{6}x-\frac{5}{6}=0$
$\Rightarrow 6x^2+\sqrt{5}x-5=0$