Question

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

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

Answer 1

Answer: (B)

$\sec A,\tan A$

Using quadratic formula. the roots of the equation $9x^2+27x+20=0$ are,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-27\pm \sqrt{{27}^2-4\times 9\times 20}}{2\times 9}$

$x=-\frac{4}{3},-\frac{5}{3}$

Given, $cosA=-\frac{3}{5}$.

Hence, $secA=\frac{1}{cosA}=-\frac{5}{3}$ and $tanA=-\sqrt{se{c}^{2}A-1}=-\frac{4}{3}$ (since A is an obtuse angle, tan A will be negative).

Thus, roots of the equation are $secA$ and $tanA$. Option B is correct.