In triangle ABC,the value of angle A is obtai | vedclass

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(C) Given $3 \cos A + 2 = 0$,we have $\cos A = -\frac{2}{3}$.Since $A$ is an angle in a triangle,$\sin A > 0$. Thus,$\sin A = \sqrt{1 - \cos^2 A} = \s

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

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(C) Given $3 \cos A + 2 = 0$,we have $\cos A = -\frac{2}{3}$.Since $A$ is an angle in a triangle,$\sin A > 0$. Thus,$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.Then,$	an A = \frac{\sin A}{\cos A} = \frac{\sqrt{5}/3}{-2/3} = -\frac{\sqrt{5}}{2}$.The quadratic equation with roots $\alpha$ and $\beta$ is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.Here,$\alpha = \sin A = \frac{\sqrt{5}}{3}$ and $\beta = 	an A = -\frac{\sqrt{5}}{2}$.Sum of roots: $\alpha + \beta = \frac{\sqrt{5}}{3} - \frac{\sqrt{5}}{2} = \frac{2\sqrt{5} - 3\sqrt{5}}{6} = -\frac{\sqrt{5}}{6}$.Product of roots: $\alpha\beta = \left(\frac{\sqrt{5}}{3}ight) \left(-\frac{\sqrt{5}}{2}ight) = -\frac{5}{6}$.The equation is $x^2 - (-\frac{\sqrt{5}}{6})x - \frac{5}{6} = 0$,which simplifies to $x^2 + \frac{\sqrt{5}}{6}x - \frac{5}{6} = 0$.Multiplying by $6$,we get $6x^2 + \sqrt{5}x - 5 = 0$.

In $\triangle ABC$,the value of $\angle A$ is obtained from the equation $3 \cos A + 2 = 0$. The quadratic equation,whose roots are $\sin A$ and $\tan A$,is

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