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

Question

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

Accepted Answer

$\cos A=-\dfrac35 \implies \sec A=-\dfrac53$. Since $A$ is obtuse, $	an A=-\sqrt{\sec^2A-1}=-\dfrac43$.

Checking: sum $=-\dfrac53-\dfrac43=-3=-\dfrac{27}9$, product $=\dfrac{20}9$, matching $9x^2+27x+20=0$. So the roots are $\sec A,	an A$.

Solutions · 1