Question

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

Answer 1

See my view of this standard.

Explanation:

Any direction can be represented by

unit vector `vec u = vec r/r`, in the direction of `vec r`.

Note that length of the vector `vec r, r >= 0`.

Here,

`vec u = cos 3.5^o veci + sin 3.5^o vec j`, or briefly,

`< cos 3.5^o, sin 3.5^o >`.

As `vecr = r < cos theta, sin theta > = < x, y >` in Cartesian frame,

this becomes

`vec u = 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]}
.

Answer 2

Explanatory notes to answer already posted.

Explanation:

By convention any angle from `0^@` to `360^@` 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^@`. However, measure of an angle or reference angle will remain same.