Question

In a triangle $ABC$, $3\cos A+2=0\Rightarrow$ quadratic equation whose roots are $\sin A,\tan A$ is:

• A. $6x^2-\sqrt{5}x-5=0$
• 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: (D)

$6x^2+\sqrt{5}x-5=0$

$cosA=\frac{-2}{3}sinA=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$

$tanA=\frac{-\sqrt{5}}{2}$

Now, in the Quadratic Equation

Sum of Roots $=\frac{\sqrt{5}}{3}+\frac{-\sqrt{5}}{2}=\frac{-\sqrt{5}}{6}$

Product of Roots $=\frac{\sqrt{5}}{3}\ast \frac{-\sqrt{5}}{2}=\frac{-5}{6}$

Therefore, Required Quadratic Equation will be

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

Therefore, Answer is $6x^2+\sqrt{5}x-5=0$