# Given that cosx=p Where x us an acute angle in degrees find in term of p i) Sin x ii)tanx iii)tan(90-x)?

Apr 29, 2018

$\sin x = \sqrt{1 - {p}^{2}}$

$\tan x = \frac{\sqrt{1 - {p}^{2}}}{p}$

$\tan \left(90 - x\right) = \frac{p}{\sqrt{1 - {p}^{2}}}$

#### Explanation:

The Pythagorean theorem states

${\sin}^{2} x + {\cos}^{2} x = 1$

Solving for $\sin x$ we have

${\sin}^{2} x = 1 - {\cos}^{2} x$

$\sin x = \sqrt{1 - {\cos}^{2} x}$

If

$\cos x = p$,

then

$\sin x = \sqrt{1 - {p}^{2}}$,

$\tan x = \sin \frac{x}{\cos} x = \frac{\sqrt{1 - {p}^{2}}}{p}$,

and

$\tan \left(90 - x\right) = \sin \frac{90 - x}{\cos} \left(90 - x\right) = \cos \frac{x}{\sin} x = \frac{p}{\sqrt{1 - {p}^{2}}}$