# If tanB=3 what is sinB?

Dec 15, 2016

Assuming you want your answer without the use of a calculator, $\sin B = 3 \cos B$

#### Explanation:

$\tan B = 3$

$\sin \frac{B}{\cos} B = 3$

$\sin B = 3 \cos B$

But $\cos B = \frac{1}{\sqrt{1 + {\tan}^{2} B}} = \frac{1}{\sqrt{10}}$

$\sin B = \frac{3}{\sqrt{10}} = \frac{3}{10} \sqrt{10}$

Dec 15, 2016

We know that $\tan \theta = \left(\text{side opposite "theta)/("side adjacent } \theta\right)$

So, the side opposite $B$ measures $3$ and the side adjacent $B$ measures $1$.

We now find the length of the hypotenuse of the imaginary triangle:

${3}^{2} + {1}^{2} = {h}^{2}$

$9 + 1 = {h}^{2}$

$h = \sqrt{10}$

We now apply the definition that $\sin \theta = \left(\text{side opposite "theta)/("hypotenuse}\right)$.

Therefore, $\sin \theta = \frac{3}{\sqrt{10}} = \frac{3 \sqrt{10}}{10}$.

Hopefully this helps!