Question

If an angle $A$ of a $△ABC$ satisfies $5\cos A+3=0,$ then the roots of the quadratic equation, $9x^2+27x+20=0$ are:

• A. $\sec A,\cot A$
• B. $\sec A,\tan A$
• C. $\tan A,\cos A$
• D. $\sin A,\sec A$

Answer 1

The correct option is B $\sec A,\tan A$

$5\cos A+3=0\Rightarrow \cos A=-\frac{3}{5}$ this means $A\in ({90}^{\circ },{180}^{\circ })$
So,$\sec A=\frac{1}{\cos A}=-\frac{5}{3}\sin A=\sqrt{1-{\cos }^{2}A}=\frac{4}{5}\Rightarrow \tan A=\frac{\sin A}{\cos A}=\frac{-4}{3}$
Now roots of the equation $9x^2+27x+20=0\Rightarrow 9x^2+12x+15x+20=03x(3x+4)+5x(3x+4)=0(3x+5)(3x+4)=0$
So, $x$= $\frac{-5}{3}\text{and}\frac{-4}{3}$
So roots are $\sec A\text{and}\tan A$.