If sin A = 1/3 and 0< A < 90, then what is cos A?

Jun 4, 2018

See below

Explanation:

If $\sin A = \frac{1}{3}$ and $0 < A < 90$

Then using ${\sin}^{2} A + {\cos}^{2} A = 1$

${\cos}^{2} A = 1 - {\sin}^{2} A = 1 - \frac{1}{9} = \frac{8}{9}$

Then $\cos A = \pm \sqrt{\frac{8}{9}} = \pm \frac{2 \sqrt{2}}{3}$

But if 0< A< 90, then the cosine is positive and $\cos A = \frac{2 \sqrt{2}}{3}$

Jun 4, 2018

$\cos A = \frac{2 \sqrt{2}}{3}$

Explanation:

Since $0 < A < {90}^{o}$ we know that $\angle A$ is in the first quadrant.

Consider the right $\triangle O A B$ where $O$ is the origin and side $A B$ is opposite $\angle A$

We are told that $\sin A = \frac{1}{3}$

$\therefore$ side $A B = 1$ and side $O A = 3$

Applying Pythagoras

${\left(O A\right)}^{2} = {\left(A B\right)}^{2} + {\left(O B\right)}^{2}$

$\therefore {3}^{2} = {1}^{2} + O {B}^{2}$

$O B = \sqrt{9 - 1}$ [Since $O B > 0$ because $\angle A$ is in the first quadrant]

$O B = \sqrt{8} = 2 \sqrt{2}$

Now, $\cos A = \frac{O B}{O A}$

$\cos A = \frac{2 \sqrt{2}}{3}$