# If sin A = 3/5, what is sec A?

Nov 24, 2014

Let $\left\{\begin{matrix}m a t h b f \left\{A\right\} \ldots \text{Adjacent" \\ mathbf{H} ... "Hypotenuse" \\ mathbf{O} ... "Opposite}\end{matrix}\right.$

Given:
$\sin A = \frac{3}{5} = \frac{m a t h b f \left\{O\right\}}{m a t h b f \left\{H\right\}} \implies \text{ Let } \left\{\begin{matrix}m a t h b f \left\{O\right\} = 3 \\ m a t h b f \left\{H\right\} = 5\end{matrix}\right.$

By Pythagorean Theorem,

$m a t h b f \left\{A\right\} = \sqrt{m a t h b f {\left\{H\right\}}^{2} - m a t h b f {\left\{O\right\}}^{2}} = \sqrt{{5}^{2} - {3}^{2}} = \sqrt{16} = 4$

Hence,

$\sec A = m a t h b f \frac{H}{m} a t h b f \left\{A\right\} = \frac{5}{4}$

I hope that this was helpful.