1. ## Lost here

I know this problem deals with Right Triangles.

When you're dealing with adjacent, aren't you dealing with cosine?

2. Yes, that's right

"?" is the adjacent and 1.3 miles is the opposite

$\displaystyle tan 57^{o}=\frac{opposite}{adjacent}$

3. Originally Posted by A Beautiful Mind
I know this problem deals with Right Triangles.

When you're dealing with adjacent, aren't you dealing with cosine?

tan is used because we don't know the hypotenuse. We know the opposite, the angle and need to find the adjacent

As $\displaystyle tan(\theta) = \frac{opposite}{adjacent}$ we use tan

4. Originally Posted by A Beautiful Mind
I know this problem deals with Right Triangles.

When you're dealing with adjacent, aren't you dealing with cosine?
Not quite.
$\displaystyle \cos \theta = \frac{\text{adj}}{{\color{red}\text{hyp}}}$
cosine is a ratio of adjacent over hypotenuse. We have a triangle where, for the given angle, the opposite side is known and the adjacent side is unknown. One of the trig functions whose ratio contains the opposite and the adjacent is
$\displaystyle \tan \theta = \frac{\text{opp}}{\text{adj}}$.
(BTW, what's the other trig function?)
So we use the tangent ratio to find the adjacent side ( = x):
\displaystyle \begin{aligned} \tan \theta &= \frac{\text{opp}}{\text{adj}} \\ \tan 57^{\circ} &= \frac{1.3}{x} \\ x\tan 57^{\circ} &= 1.3 \\ x &= \frac{1.3}{\tan 57^{\circ}} \end{aligned}

01

EDIT: beaten to it again!