# FIND REFERENCE ANGLE

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.

## Given Angle is Positive

If θ is positive but greater than 360°, find the positive angle between 0° and 360° that is coterminal with θ°.

To get the coterminal angle, divide θ by 360° and take the remainder.

## Given Angle is Negative

If θ is negative, add multiples of 360° to θ make the angle as positive such that it is between 0° and 360°.

Once we have the given angle as positive and also it is between 0° and 360°, easily we can find the reference angle as explained below.

Find the reference angle.

Problem 1 :

Solution :

The given angle -230º is negative.

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

-230º + 360º = 130º

130º is positive and less than 360º.

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

So, the reference angle is

= 180º  - 130º

= 50º

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 0º 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 0º 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 0º 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 :

Solution :

The given angle lies in the IIIrd quadrant. Writing the given angle as multiple of 360, we get

Terminal angle = 4π - (31π/9)

= (36 - 31)π/9

= 5π/9

Since the terminal angle lies in 3rd quadrant, to find reference angle, we use

Reference angle = π - (5π/9 )

= (9π - 5π)/9

= 4π/9

Problem 6 :

Solution :

The given angle lies in the IVth quadrant. Writing the given angle as multiple of 360, we get

Terminal angle = 720 - 640

= 80

So, the reference angle is 80.

Problem 7 :

-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 8 :

-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 2nd quadrant, so the reference angle will be

= π - (11π/12)

π/12

Problem 8 :

-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 2nd quadrant, so the reference angle will be

= π - (17π/18)

π/18

Problem 10 :

-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

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