# How to solve this problem?

Aug 14, 2017

#### Explanation:

I would use identities.

We know that ${\sin}^{2} x + {\cos}^{2} x = 1$, then we have:

${\left(\frac{5}{13}\right)}^{2} + {\cos}^{2} x = 1$

${\cos}^{2} x = 1 - \frac{25}{169}$

${\cos}^{2} x = \frac{144}{169}$

$\cos x = \pm \frac{12}{13}$

However, we know the answer must be negative because of the C-A-S-T rule, which is shown in the following picture.

So only sine is positive on $\frac{\pi}{2} < x < \pi$ , which is equivalent to 90˚ < x <180˚. So cosine will be $- \frac{12}{13}$.

Now we can use the quotient identity, which states that $\tan \theta = \sin \frac{\theta}{\cos} \theta$. This stems from a right triangle with sides $x$ and $y$ and hypotenuse $r$. If $\theta$ is adjacent $x$ and opposite $y$, then the ratios are

$\cos \theta = \frac{x}{r}$
$\sin \theta = \frac{y}{r}$
$\tan \theta = \frac{y}{x}$

Now notice that

$\frac{\frac{y}{r}}{\frac{x}{r}} = \frac{y}{x}$ or $\sin \frac{\theta}{\cos} \theta = \tan \theta$

$\tan \theta = \sin \frac{\theta}{\cos} \theta = \frac{\frac{5}{13}}{- \frac{12}{13}} = - \frac{5}{12}$

Hopefully this helps!

Aug 14, 2017

$\text{d)}$ $- \frac{5}{12}$

#### Explanation:

We know that $\sin \left(\theta\right) = \frac{5}{13}$.

$\sin \left(\theta\right)$ is also equal to frac("opposite")("hypotenuse").

$R i g h t a r r o w \frac{\text{opposite")("hypotenuse}}{=} \frac{5}{13}$

So, the opposite and hypotenuse are equal to $5$ and $13$, or they can be multiples of them.

For this problem, we can simply consider the opposite and hypotenuse to be $5$ and $13$, respectively.

The hypotenuse of a triangle is its longest side.

Using Pythagoras' theorem:

$R i g h t a r r o w {\text{opposite"^(2) + "adjacent"^(2) = "hypotenuse}}^{2}$

$R i g h t a r r o w {5}^{2} + {\text{adjacent}}^{2} = {13}^{2}$

$R i g h t a r r o w {\text{adjacent}}^{2} = {13}^{2} - {5}^{2}$

$R i g h t a r r o w {\text{adjacent}}^{2} = 169 - 25$

$R i g h t a r r o w {\text{adjacent}}^{2} = 144$

$R i g h t a r r o w \sqrt{{\text{adjacent}}^{2}} = \pm \sqrt{144}$

$\therefore \text{adjacent} = \pm 12$

Now, $\tan \left(\theta\right)$ is equal to frac("opposite")("adjacent"):

$R i g h t a r r o w \tan \left(\theta\right) = \frac{5}{\pm 12}$

The interval that we are provided with is $\frac{\pi}{2} < \theta < \pi$, i.e. the second quadrant.

In the second quadrant, all values of $\tan \left(\theta\right)$ are negative.

So, $\tan \left(\theta\right)$ cannot be equal to $\frac{5}{+ 12} = \frac{5}{12}$.

Therefore, $\tan \left(\theta\right)$ is equal to $\frac{5}{- 12} = - \frac{5}{12}$.

In conclusion, the final answer is $\text{d)}$ $- \frac{5}{12}$.