# How to Solve Coterminal Angles and Reference Angles? (+FREE Worksheet!)

If you want to learn how to solve Coterminal angles and Reference angles problems, you are in the right place.

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## Step by step guide to solve Coterminal Angles and Reference Angles Problems

- Coterminal angles are equal angles.
- To find a coterminal of an angle, add or subtract \(360\) degrees (or \(2π\) for radians) to the given angle.
- Reference angle is the smallest angle that you can make from the terminal side of an angle with the \(x\)-axis.

### Coterminal Angles and Reference Angles – Example 1:

Find positive and negative coterminal angles to angle \(65^\circ\).

**Solution**:

\(65^\circ-360^\circ=-295^\circ \)

\( 65^\circ+360^\circ=425^\circ \)

\( -295^\circ\) and a \(425^\circ\) are coterminal with a \(65^\circ\).

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### Coterminal Angles and Reference Angles – Example 2:

Find a positive and negative coterminal angle to angle \(\frac{π}{2}\).

**Solution**:

\(\frac{π}{2}+2π=\frac {π+(2 ×2π) } {2} =\frac {π+4 π }{2}= \frac{5π}{2} \)

\( \frac{π}{2}-2π= \frac {π-(2 ×2π) } {2} =\frac {π-4 π }{2}= -\frac{3π}{2 }\)

### Coterminal Angles and Reference Angles – Example 3:

Find positive and negative coterminal angles to angle \(70^\circ\).

**Solution**:

\(70^\circ-360^\circ=-290^\circ \)

\(70^\circ+360^\circ=430^\circ \)

\( -290^\circ\) and a \(430^\circ\) are coterminal with a \(70^\circ\).

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### Coterminal Angles and Reference Angles – Example 4:

Find positive and negative coterminal angles to angle \(\frac{π}{4}\).

**Solution:**

\(\frac{π}{4}+2π= \frac {π+(4 ×2π) } {4} =\frac {π+8π }{4} =\frac{9π}{4 }\)

\( \frac{π}{4}-2π= \frac {π-(4 ×2π) } {4} =\frac {π-8 π }{4} =-\frac{7π}{4 }\)

## Exercises for Solving Coterminal Angles and Reference Angles

### Find a coterminal angle between \(0\) and \(2π\) for each given angle.

- \(\color{blue}{\frac{14π}{5}=} \\ \)
- \(\color{blue}{-\frac{16π}{9}=} \\ \)
- \(\color{blue}{\frac{41π}{18}=} \\ \)
- \(\color{blue}{-\frac{19π}{12}=} \)

- \(\color{blue}{\frac{4π}{5}} \\ \)
- \(\color{blue}{\frac{2π}{9}} \\ \)
- \(\color{blue}{\frac{5π}{18}} \\ \)
- \(\color{blue}{\frac{5π}{12}}\)

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