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In $\triangle A B C$, 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
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Verified Answer
The correct answer is:
$6 x^2+\sqrt{5} x-5=0$
We have,
$$
\begin{array}{rlrl}
3 \cos A+2 & =0 \\
\Rightarrow & \cos A & =\frac{-2}{3} \\
\therefore & & \sin A=\sqrt{1-\cos ^2 A} & =\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \\
& \text { and } \quad & \tan A=\frac{\sin A}{\cos A} & =\frac{\sqrt{5} / 3}{-2 / 3}=\frac{-\sqrt{5}}{2}
\end{array}
$$
Since, $\sin A$ and $\tan A$ are the roots of required equation.
Hence, equation can be written as
$$
\begin{aligned}
& x^2-(\sin A+\tan A) x+\sin A \tan A=0 \\
& \Rightarrow x^2-\left(\frac{\sqrt{5}}{3}-\frac{\sqrt{5}}{2}\right) x+\left(\frac{\sqrt{5}}{3}\right)\left(\frac{-\sqrt{5}}{2}\right)=0 \\
& \Rightarrow \quad x^2+\frac{\sqrt{5}}{6} x-\frac{5}{6}=0 \\
& \Rightarrow \quad 6 x^2+\sqrt{5} x-5=0
\end{aligned}
$$
$$
\begin{array}{rlrl}
3 \cos A+2 & =0 \\
\Rightarrow & \cos A & =\frac{-2}{3} \\
\therefore & & \sin A=\sqrt{1-\cos ^2 A} & =\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \\
& \text { and } \quad & \tan A=\frac{\sin A}{\cos A} & =\frac{\sqrt{5} / 3}{-2 / 3}=\frac{-\sqrt{5}}{2}
\end{array}
$$
Since, $\sin A$ and $\tan A$ are the roots of required equation.
Hence, equation can be written as
$$
\begin{aligned}
& x^2-(\sin A+\tan A) x+\sin A \tan A=0 \\
& \Rightarrow x^2-\left(\frac{\sqrt{5}}{3}-\frac{\sqrt{5}}{2}\right) x+\left(\frac{\sqrt{5}}{3}\right)\left(\frac{-\sqrt{5}}{2}\right)=0 \\
& \Rightarrow \quad x^2+\frac{\sqrt{5}}{6} x-\frac{5}{6}=0 \\
& \Rightarrow \quad 6 x^2+\sqrt{5} x-5=0
\end{aligned}
$$
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