# How do you find the reference angle of theta=3.5 and sketch the angle in standard position?

Jul 30, 2018

See my view of this standard.

#### Explanation:

Any direction can be represented by

unit vector $\vec{u} = \frac{\vec{r}}{r}$, in the direction of $\vec{r}$.

Note that length of the vector $\vec{r} , r \ge 0$.

Here,

$\vec{u} = \cos {3.5}^{o} \vec{i} + \sin {3.5}^{o} \vec{j}$, or briefly,

$< \cos {3.5}^{o} , \sin {3.5}^{o} >$.

As $\vec{r} = r < \cos \theta , \sin \theta > = < x , y >$ in Cartesian frame,

this becomes

$\vec{u} = \frac{1}{\sqrt{{x}^{2} + {y}^{2}}} < x , y >$

See graph of $< \cos {3.5}^{o} , \sin {3.5}^{o} > =$<0.9991, 0.0610 >#,

nearly, using befitting domain and range, and scaling, for better

visual effect.

graph{ 0.9991y-0.0601 x= 0[0 0.9991 -0.050 0.5]}
.

Jul 31, 2018

By convention any angle from ${0}^{\circ}$ to ${360}^{\circ}$ is measured from the positive $x$-axis to the line that ends the angle, also called the terminal side.
Reference angle is taken as angle formed by the $x$-axis and the terminal side. In the figure below, one can see all the reference angles possible for each quadrant.
We are so much used to seeing east as positive $x$-axis that we forget that selection of this axis is entirely optional. One can select north as positive $x$-axis! Consequently, the figures will get rotated by ${90}^{\circ}$. However, measure of an angle or reference angle will remain same.