Question

If ABCD is a cyclic quadrilateral such that $12\tan A-5$ and $5\cos B+3=0$ then the quadratic equation whose roots are $\cos C,\tan D$ is

• A. $39x^2-16x-48=0$
• B. $39x^2+88x-48=0$
• C. $39x^2-88x+48=0$
• D. none of these

Answer 1

Answer: (A)

$39x^2-16x-48=0$

$\tan A=\frac{5}{12}$ and $\cos B=\frac{-3}{5}$

For cyclic quadrilateral, $A+C=\pi$ and $B+D=\pi$

Therefore, $\tan C=\tan (\pi -A)=\frac{-5}{12}$ and $\cos D=\cos (\pi -C)=\frac{3}{5}$

$\Rightarrow \cos C=\frac{-12}{13}$ and $\tan D=\frac{4}{3}$

Hence required quadratic equation is,

$x^2-(\frac{-12}{13}+\frac{4}{3})+(\frac{-12}{13}.\frac{4}{3})=0$

$\Rightarrow 39x^2-16x-48=0$