Question

In the given figure $△ABC$ is a right-angled triangle, where $\angle B={90}^o,\angle C={50}^o$. Write the ratios of $\sin {50}^o$ and $\cos {50}^o$.

Answer 1

In $△ABC$, we have
$\cos {50}^o=$ adjacent side/hypotenuse$=\frac{BC}{AC}$
$\sin {50}^o=$ opposite/hypotenuse$=\frac{AB}{AC}$
Therefore, $\frac{\sin {50}^o}{\cos {50}^o}=\frac{\left(\frac{AB}{AC}\right)}{\left(\frac{BC}{AC}\right)}=\frac{AB}{BC}$