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

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(B) Given $5\cos A + 3 = 0$,so $\cos A = -\frac{3}{5}$.Since $A$ is an angle of a triangle,$\sin A$ must be positive. Thus,$\sin A = \sqrt{1 - \cos^2

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$15x^2 + 8x - 16 = 0$

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(B) Given $5\cos A + 3 = 0$,so $\cos A = -\frac{3}{5}$.Since $A$ is an angle of a triangle,$\sin A$ must be positive. Thus,$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - (-\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.Then,$	an A = \frac{\sin A}{\cos A} = \frac{4/5}{-3/5} = -\frac{4}{3}$.Let the roots be $\alpha = \sin A = \frac{4}{5}$ and $\beta = 	an A = -\frac{4}{3}$.Sum of roots: $\alpha + \beta = \frac{4}{5} - \frac{4}{3} = \frac{12 - 20}{15} = -\frac{8}{15}$.Product of roots: $\alpha \cdot \beta = (\frac{4}{5})(-\frac{4}{3}) = -\frac{16}{15}$.The quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - (-\frac{8}{15})x + (-\frac{16}{15}) = 0$.$x^2 + \frac{8}{15}x - \frac{16}{15} = 0$.Multiplying by $15$,we get $15x^2 + 8x - 16 = 0$.

In a triangle $ABC$,the value of $\angle A$ is given by $5\cos A + 3 = 0$. Then,the equation whose roots are $\sin A$ and $\tan A$ is:

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