Originally Posted by

**skeeter** cosine of any angle must be between -1 and 1 ... it cannot equal 5.

$\displaystyle B$ is in quadrant II

correction ... messed up my signs

$\displaystyle \tan{B} = -\frac{\sqrt{11}}{5}$

$\displaystyle \tan{B} = \frac{\sqrt{11}}{-5} = \frac{opposite \, side}{adjacent \, side}$

$\displaystyle (\sqrt{11})^2 + (-5)^2 = (hypotenuse)^2$

$\displaystyle 11 + 25 = 36$

$\displaystyle hypotenuse = 6$

$\displaystyle \cos{B} = \frac{adjacent \, side}{hypotenuse} = -\frac{5}{6}$

or, using an identity ...

$\displaystyle \tan{B} = -\frac{\sqrt{11}}{5}

$

$\displaystyle \sec^2{B} = 1 + \tan^2{B} $

$\displaystyle \sec^2{B} = 1 + \frac{11}{25} = \frac{36}{25}$

$\displaystyle \sec{B} = -\frac{6}{5}$

$\displaystyle \cos{B} = -\frac{5}{6} $