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

2 Answers
Jul 30, 2018

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]}
.

Jul 31, 2018

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.

tutorvista.com

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.