What is cosΘ if tanΘ = 6/17?

Apr 1, 2018

$\implies \frac{17 \sqrt{13}}{65}$

Explanation:

We know $\tan \theta = \text{opposite"/"adjacent}$

Hence opposite" = 6  and $\text{adjacent} = 17$

Hence using Pythagorus rule: ${a}^{2} + {b}^{2} = {c}^{2}$

$\implies {6}^{2} + {17}^{2} = {c}^{2} \implies c = 5 \sqrt{13}$

$\implies \text{hypotenuse} = 5 \sqrt{13}$

We know $\cos \theta = \text{adjacent" / "hypotenuse}$

$\implies \cos \theta = \frac{17}{5 \sqrt{13}}$

$\implies \frac{17 \sqrt{13}}{65}$

Apr 1, 2018

$\textcolor{b l u e}{\frac{17 \sqrt{13}}{65}}$

Explanation: From diagram, we have:

$\tan \left(\theta\right) = \text{opposite"/"adjacent} = \frac{6}{17}$

$\cos \left(\theta\right) = \text{adjacent"/"hypotenuse} = \frac{17}{c}$

We need to find $\boldsymbol{c}$:

Using Pythagoras' Theorem:

${c}^{2} = {6}^{2} + {17}^{2}$

$c = \sqrt{{\left(6\right)}^{2} + {\left(17\right)}^{2}} = \sqrt{325} = 5 \sqrt{13}$

$\cos \left(\theta\right) = \text{adjacent"/"hypotenuse} = \frac{17}{5 \sqrt{13}} = \textcolor{b l u e}{\frac{17 \sqrt{13}}{65}}$