Question

If the $\angle A$ of a triangle $ABC$ is given by the equation $5\cos A+3=0$, then $\sin A$ and $\tan A$ are the roots of the equation

• A. $15x^2-8x-16=0$
• B. $15x^2-8\sqrt{2x}+16=0$
• C. $15x^2-8x+16=0$
• D. $15x^2+8x-16=0$

Answer 1

Answer: (D)

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

$5\cos A+3=0$

$5\cos A=-3$

$\cos A=-\frac{3}{5}$

$\cos A$ is negative, so angle $A$ is an obtuse angle.

$\angle A>{90}^{\circ }$

$\sin A=\sqrt{1-{\cos }^{2}A}$

$\sin A=\sqrt{1-{\left(\frac{-3}{5}\right)}^2}$

$\sin A=\sqrt{\frac{16}{25}}=\frac{4}{5}$ ( $\sin A$ will be positive because angle $A$ lies in second quadrant )

$\tan A=\frac{\sin A}{\cos A}=-\frac{4}{3}$

If $\sin A$ and $\tan A$ are the roots of a quadratic equation, then the equation is,

$x^2-x(\sin A+\tan A)+(\sin A\cdot \tan A)=0$

$x^2-x(\frac{4}{5}-\frac{4}{3})+(\frac{4}{5}\times \frac{-4}{3})=0$

$x^2-x\left(\frac{-8}{15}\right)-\left(\frac{16}{15}\right)=0$

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