# Question #cca9c

Apr 27, 2017

$x = {143}^{o}$

#### Explanation:

First of all, it helps to know that Sin (${37}^{o}$) = $\frac{3}{5}$ and Cos (${37}^{o}$) = $\frac{4}{5}$.

Now according to your question, Sin x is positive and cos x is negative. This is only possible in the second quadrant of the cartesian plane, where Sin x is positive, and Cos x & Tan x are negative.

In the second quadrant, all angles lie between ${90}^{o} \mathmr{and} {180}^{o}$.

So we know that $S \in \left(x\right) = S \in \left({180}^{o} - x\right)$ and $- C o s \left(x\right) = C o s \left({180}^{o} - x\right)$

Putting x as ${37}^{o}$, we get

$S \in \left({37}^{o}\right) = S \in \left({143}^{o}\right)$
$S \in \left({143}^{o}\right) = \frac{3}{5}$

Also,

$- C o s \left({37}^{o}\right) = C o s \left({143}^{o}\right)$
$C o s \left({143}^{o}\right) = - \frac{4}{5}$

As all conditions of the question are satisfied, $x = {143}^{o}$

Apr 27, 2017

$x \approx {143.13}^{\circ}$ or 2.498 radians

#### Explanation:

We know that x is in the second quadrant, because the sine is positive and cosine is negative. If we use the cosine function and take its inverse, the calculator will do all of the work to give us the correct angle.

$\cos \left(x\right) = - \frac{4}{5}$

Use the inverse cosine function on both sides:

${\cos}^{-} 1 \left(\cos \left(x\right)\right) = {\cos}^{-} 1 \left(- \frac{4}{5}\right)$

$x = {\cos}^{-} 1 \left(- \frac{4}{5}\right)$

$x \approx {143.13}^{\circ}$ or 2.498 radians