Question

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

• A. $secA,cotA$
• B. $sinA,secA$
• C. $secA,tanA$
• D. $tanA,cosA$

Answer 1

The correct option is C $secA,tanA$

Given If an angle a of abc satisfies $5cosA+3=0$,

then the roots of the quadratic equation $9x^2+27x+20=0$ are

Eqn is $5cosA+3=0$

$⟹5cosA=-3⟹cosA=-\frac{3}{5}$

So $secA=\frac{1}{cosA}⟹secA=\frac{-1}{\frac{-3}{5}}⟹secA=-\frac{5}{3}$

$se{c}^{2}A-ta{n}^{2}A=1⟹ta{n}^{2}A=se{c}^{2}A-1⟹tanA=\sqrt{se{c}^{2}a-1}=\sqrt{\frac{25}{9}-1}=\sqrt{\frac{16}{9}}$

$tanA=\frac{4}{3}$

So roots of the equation are $secA,tanA$