FINDING REFERENCE ANGLES FOR NEGATIVE ANGLES

Let A be an angle in standard position. The reference angle B associated with A is the acute angle formed by the terminal side of A and the x-axis.

Ensure that the given angle is positive and it is between 0° and 360°.

What if the given angle does not meet the criteria above :

Let θ be the angle given.

Find the reference angle.

Problem 1 :

Solution :

The given angle -230^º is negative.

Add multiples of 360º to -230º to make the angle as positive such that it is between 0^º and 360^º.

-230^º + 360^º = 130^º

130^º is positive and less than 360^º.

The terminal side of the angle 130^º is in quadrant II, as shown above.

So, the reference angle is

= 180^º  - 130^º

= 50^º

Problem 2 :

Solution :

The given angle -25π/18 is negative.

Add multiples of 360º to -25π/18 to make the angle as positive such that it is between 0^º and 2π.

-25π/18 + 2π = 11π/18

11π/18 is positive and less than 2π.

The terminal side of the angle  11π/18 is in quadrant II, as shown above.

So, the reference angle is

= π - 11π/18

= 7π/18

Problem 3 :

Solution :

The given angle -7π/9 is negative.

Add multiples of 2π to -7π/9 to make the angle as positive such that it is between 0^º and 2π.

-7π/9 + 2π = 11π/9

11π/9 is positive and less than 2π.

The terminal side of the angle 11π/9 is in quadrant III, as shown above.

So, the reference angle is

= 11π/9 - π

= 2π/9

Problem 4 :

Solution :

The given angle -29π/18 is negative.

Add multiples of 2π to -29π/18 to make the angle as positive such that it is between 0^º and 2π.

-29π/18 + 2π = (-29π + 36π)/18

= 7π/18

The same is taken as reference angle since the angle lies in the first quadrant.

Problem 5 :

-510º

Solution :

Since the given angle is negative, to get positive angle, we write as combination of 360.

720 - 510 = 210º

The terminal angle 210º lies is 3rd quadrant, so the reference angle will be 

= 210 - 180

= 30º

Problem 6 :

-13π/12

Solution :

Since the given angle is negative, to get positive angle, we write as combination of 2π.

= -13π/12 + 2π

= 11π/12

The terminal angle 11π/12 lies is 2^nd quadrant, so the reference angle will be 

= π - (11π/12)

= π/12

Problem 7 :

-19π/18

Solution :

Since the given angle is negative, to get positive angle, we write as combination of 2π.

= -19π/18 + 2π

= 17π/18

The terminal angle 17π/18 lies is 2^nd quadrant, so the reference angle will be 

= π - (17π/18)

= π/18

Problem 8 :

-250º

Solution :

Since the given angle is negative, to get positive angle, we write as combination of 360.

360 - 250 = 110º

The terminal angle 110º lies is 2nd quadrant, so the reference angle will be 

= 180 - 110

= 70º

Problem 9 :

-170º

Solution :

Since the given angle is negative, to get positive angle, we write as combination of 360.

-170º + 360º = 190º

The terminal angle 190º lies is 3rd quadrant, so the reference angle will be 

= Given angle - 180º

= 190º - 180º

= 10º

Problem 10 :

-400º

Solution :

Since the given angle is negative, to get positive angle, we write as combination of 360.

-400º + 360º = -40º

-40º + 360º = 320º

The terminal angle 320º lies is 4th quadrant, so the reference angle will be 

= 360º - given angle

= 360º - 320º

= 40º

Problem 11 :

-5π/3

Solution :

Since the given angle is negative, to get positive angle, we write as combination of 2π.

= -5π/3 + 2π

= π/3

The terminal angle π/3 lies is 1^st quadrant, so the reference angle will be the same.

= π/3