# Question #f6d94

Apr 8, 2017

$\cos \left(b\right) = \sqrt{\frac{289}{338}} \approx 0.924680985 \left(\text{radians}\right)$

#### Explanation:

Here are 2 ways you could do this.

Method 1: ...almost feels like cheating ;)
If $\tan \left(2 b\right) = \frac{119}{120}$
$\textcolor{w h i t e}{\text{XXX}} 2 b = \arctan \left(\frac{119}{120}\right) \approx 0.7812140874$
which implies (after dividing by 2)
$\textcolor{w h i t e}{\text{XXX}} b = 0.3906070437$
Then
use your calculator again to find
$\textcolor{w h i t e}{\text{XXX}} \cos \left(b\right) = \cos \left(0.3906070437\right) \approx 0.9246780985$

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Method 2: possibly more virtuous (???)
If $\tan \left(2 b\right) = \frac{119}{120}$
then we can think of angle $\left(2 b\right)$ as being the angle of a right triangle in standard position with the horizontal ($x$) component equal to $120$ and the vertical ($y$) component equal to $119$.
The hypotenuse would be equal to $\sqrt{{120}^{2} + {119}^{2}} = 169$
(and, yes, I did use a calculator to discover this).

$\cos \left(2 b\right) = \left(\text{horizontal")/("hypotenuse}\right) = \frac{120}{169}$

Then, if we remember the double angle formula:
$\textcolor{w h i t e}{\text{XXX}} \cos \left(2 b\right) = 2 {\cos}^{2} \left(b\right) - 1$
we can re-arrange the terms to get
$\textcolor{w h i t e}{\text{XXX}} \cos \left(b\right) = \sqrt{\frac{\cos \left(2 b\right) + 1}{2}}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \sqrt{\frac{\frac{120}{169} + 1}{2}}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \sqrt{\frac{\frac{289}{169}}{2}}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \sqrt{\frac{289}{338}}$

and with the aid of my calculator (again)
$\textcolor{w h i t e}{\text{XXXXXX}} \approx 0.924680985$