Find Reference Angle

[harpazo]

Find the reference angle for A and B.

A. 35°

Let R = reference angle

In quadrant 1, R = given angle.

So, the reference angle of 35° is 35°. I want to know why this is the case.

B. 250°

I will use R = given angle - 180°.

R = 250° - 180°

R = 70°

The reference angle is 70°.

How is this done for angles given in radians?

Example: Find the reference angle for 5pi/4.

 

[topsquark]

35 degrees is its own reference angle by definition. Nothing confusing about.

You do the reference angles the same way no matter what the unit. For radians note that you want an angle between 0 and [math]\pi / 2[/math] rad. The easiest way to learn it is to remember that
[math]\begin{cases} 0^o = 0 \text{ rad} \\ 90^o = \dfrac{ \pi }{2} \text{ rad} \\ 180^o = \pi \text{ rad} \\ 270^o = \dfrac{3 \pi }{2} \text{ rad} \\ 360^o = 2 \pi \text{ rad} \end{cases}[/math]
-Dan

 

[harpazo]

35 degrees is its own reference angle by definition. Nothing confusing about.

You do the reference angles the same way no matter what the unit. For radians note that you want an angle between 0 and [math]\pi / 2[/math] rad. The easiest way to learn it is to remember that
[math]\begin{cases} 0^o = 0 \text{ rad} \\ 90^o = \dfrac{ \pi }{2} \text{ rad} \\ 180^o = \pi \text{ rad} \\ 270^o = \dfrac{3 \pi }{2} \text{ rad} \\ 360^o = 2 \pi \text{ rad} \end{cases}[/math]
-Dan

Thank you for the list of angles. Back in 2006, Soroban suggested for me to use the following in terms of finding reference angles:

Let R = reference angle

Let Q = quadrant

Degrees

In Q 1, R = angle in Q 1
In Q 2, R = 180° - angle in Q 2
In Q 3, R = angle in Q 3 - 180°
In Q 4, R = 360° - angle in Q 4

Radians

In Q 1, R = angle in Q 1 (in rad).
In Q 2, R = pi - angle in Q 2 (in rad).
In Q 3, R = angle in Q 3 (in rad) - pi
In Q 4, R = 2pi - angle in Q 4 (in rad).

By the way, where is Soroban? I recall that he was 75 back in 2006. Is he alive? What is his real name? I miss him big time. He is 89? 90?