# Find the exact value of cos(x) if tan(x)=5 and is in quadrant I?

Apr 7, 2018

$\cos \left(x\right) = \frac{1}{\sqrt{26}}$

#### Explanation:

This problem is solved using the Pythagorean Theorem and knowledge of "soh-cah-toa".

Recall that "soh-cah-toa" reminds us that $\sin \left(x\right) = \text{opposite"/"hypotenuse}$, $\cos \left(x\right) = \text{adjacent"/"hypotenuse}$ and $\tan \left(x\right) = \text{opposite"/"adjacent}$, where 'opposite', 'adjacent' and 'hypotenuse' refer to sides of a right triangle.

Our equality $\tan \left(x\right) = 5$ refers to a right triangle where the side opposite from angle $x$ is 5 times larger than the side adjacent to $x$. The following is one such triangle (not drawn to scale).

Using the Pythagorean Theorem, we can find the length of the hypotenuse.

${a}^{2} + {b}^{2} = {c}^{2}$,
${1}^{2} + {5}^{2} = {c}^{2}$,
$26 = {c}^{2}$,
$c = \sqrt{26}$.

Then $\cos \left(x\right)$ refers to our same triangle. See that $\cos \left(x\right) = \text{adjacent"/"hypotenuse} = \frac{1}{\sqrt{26}}$.

Note: This triangle extends up and to the right, into the first quadrant.