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.
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.
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.
Angles in quadrants 1st quadrant 2nd quadrant 3rd quadrant 4th quadrant |
Formula the same 180 - given angle given angle - 180 360 - given angle |
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 :
Solution :
The given angle lies in the III^{rd} 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 IV^{th} 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 2^{nd} 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 2^{nd} 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
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM