# If sinB=4/9 what is tanB?

$\pm \frac{4 \sqrt{65}}{65}$

#### Explanation:

$\sin B = \text{opp"/"hyp} = \frac{4}{9}$

We can find the adjacent via the Pythagorean Theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

${4}^{2} + {b}^{2} = {9}^{2}$

$16 + {b}^{2} = 81$

${b}^{2} = 65$

$b = \sqrt{65} \implies 8 < b < 9$

Now that we have lengths, let's now talk about the position of the reference angle and triangle. Sine is positive in Q1 and Q4. In Q1, tangent is positive but in Q4, it is negative. And so the final answer is:

$\tan B = \text{opp"/"adj} = \frac{\pm 4}{\sqrt{65}} = \frac{\pm 4}{\sqrt{65}} \left(1\right) = \frac{\pm 4}{\sqrt{65}} \left(\frac{\sqrt{65}}{\sqrt{65}}\right) = \pm \frac{4 \sqrt{65}}{65}$