# If sin theta = 3/5 and theta is acute, how do you calculate cos theta and tan theta?

Jul 26, 2016

$\cos \theta = \frac{4}{5}$
$\tan \theta = \frac{3}{4}$

#### Explanation:

$\sin \theta = \frac{3}{5}$
costheta=sqrt(1-sin^2theta
or
$\cos \theta = \sqrt{1 - {\left(\frac{3}{5}\right)}^{2}}$
or
costheta=sqrt(1-9/25
or
$\cos \theta = \sqrt{\frac{25 - 9}{25}}$
or
$\cos \theta = \sqrt{\frac{16}{25}}$
or
$\cos \theta = \frac{4}{5}$------------Ans $1$

$\tan \theta = \sin \frac{\theta}{\cos} \theta$
or
$\tan \theta = \frac{\frac{3}{5}}{\frac{4}{5}}$
or
$\tan \theta = \frac{3}{4}$-------------Ans $2$

Jul 26, 2016

$\cos \theta = \frac{4}{5} , \mathmr{and} , \tan \theta = \frac{3}{4}$.

#### Explanation:

We will use the Identity $: {\sin}^{2} \theta + {\cos}^{2} \theta = 1$.

$\therefore {\cos}^{2} \theta = 1 - {\sin}^{2} \theta = 1 - {\left(\frac{3}{5}\right)}^{2} = 1 - \frac{9}{25} = \frac{16}{25}$

$\therefore \cos \theta = \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$

But we know that theta is acute , so, it is in the first quadrant, in which, all the Trgo. ratios are $+ v e$

Hence, $\cos \theta = + \frac{4}{5}$.

Finally, $\tan \theta = \sin \frac{\theta}{\cos} \theta$..............[Defn.]

$= \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$