Question

If $\sin A=\frac{5}{13}$ then evaluate $\cos A$ and $\tan A$.

Answer 1

Given: $\sin A=\frac{5}{13}$

$\cos A=\sqrt{1-{\sin }^{2}A}$ [$\because {\sin }^{2}A+{\cos }^{2}A=1$]

$=\sqrt{1-(\frac{5}{13}{)}^2}$

$=\sqrt{1-\left(\frac{25}{169}\right)}$

$=\sqrt{\left(\frac{169-25}{169}\right)}$

$=\sqrt{\left(\frac{144}{169}\right)}$

$=\sqrt{\left(\frac{12\times 12}{13\times 13}\right)}$

$\therefore \cos A=\pm \frac{12}{13}$

$\tan A=\frac{\sin A}{\cos A}=\frac{\left(\frac{5}{13}\right)}{(\pm \frac{12}{13})}=\pm \frac{5}{13}\times \frac{13}{12}=\pm \frac{5}{12}$