Question

In $\Delta ABC$ value of A obtain from equation 3 cos A+2=0, then the equation whose roots are sin A and tan A is

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

Answer 1

The correct option is A $6x^2+\sqrt{5}x-5=0$

$3\cos A+2=0$

$\Rightarrow \cos A=-\frac{2}{3}$

Since, cos A is negative.

i.e. $\frac{\pi }{2}<\angle A<\pi$

$\sin A=\pm \sqrt{1-{\cos }^{2}A}({\sin }^{2}A+{\cos }^{2}A=1)$

$=\pm \sqrt{1-\frac{4}{9}}$

$=+\sqrt{\frac{5}{9}}$

$\text{angle}Aisin1stquadrant$

$-\frac{\sqrt{5}}{3}$

$\tan A=\frac{\sin A}{\cos A}$

$=\frac{\sqrt{5}}{3}\times \frac{3}{-2}=\frac{\sqrt{5}}{-2}$

$

$\text{Equation whose roots are}\sin A\text{and}$

$\text{tam}A\text{:}$

$x^2-(\sin A+\tan A)+\sin A\tan A=0$

$(x^2-(\alpha +\beta )+\alpha \beta =0\text{, where}\alpha \text{and}\beta )$

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

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