# How do you sketch the angle whose terminal side in standard position passes through (6,8) and how do you find sin and cos?

Aug 18, 2017

#### Explanation:

To sketch the angle in standard position,

One side of the angle is the positive $x$ axis (the right side of the horizontal axis). That is the initial side.

The terminal side has one end at the origin (the point $\left(0 , 0\right)$, also the intersection of the two axes) and goes through the point $\left(6 , 8\right)$. So to locate the point $\left(6 , 8\right)$, starting at the origin and count $6$ to the right and up $8$. That will get you to the point $\left(6 , 8\right)$. Put a dot there.
Now draw a line from the origin through the point $\left(6 , 8\right)$. Your sketch should look a lot like this:

If we knew how the angle was made (which direction and how many times around the circle), we would show that also.

Give the angle a name. I will use $\theta$ (that is the Greek letter "theta")

Memorize this

If the point $\left(a , b\right)$ lies on the terminal side of an angle in standard position, then let $r = \sqrt{{a}^{2} + {b}^{2}}$

The angle has sine $\frac{b}{r}$ and it has cosine $\frac{a}{r}$

For this question

We have $\left(a , b\right) = \left(6 , 8\right)$, so

$r = \sqrt{{\left(6\right)}^{2} + {\left(8\right)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$.

So the sine of $\theta$ is

$\sin \left(\theta\right) = \frac{8}{10}$

When we think we have finished, we should make sure that fracions are reduced.

$\frac{8}{10} = \frac{2 \cdot 4}{2 \cdot 5} = \frac{\cancel{2} \cdot 4}{\cancel{2} \cdot 5} = \frac{4}{5}$

$\sin \theta = \frac{4}{5}$
The cosine of $\theta$ is $\frac{a}{r}$ so we have
$\cos \left(\theta\right) = \frac{6}{10} = \frac{2 \cdot 3}{2 \cdot 5} = \frac{\cancel{2} \cdot 3}{\cancel{2} \cdot 5} = \frac{3}{5}$
$\cos \left(\theta\right) = \frac{3}{5}$