Question

13.6 complementary and supplementary angles (dok 1, 2)
two angles are complementary if the sum of their angles measures 90°. complementary and supplementary angles may be adjacent but do not need to be.
a linear pair is a pair of adjacent angles that are supplementary. below, the angles 32° and 148° are a linear pair, as are the angles 70° and 110°.
complementary angles
supplementary angles
calculate the measure of each unknown angle. (dok 2)

1. (mangle a=)
2. (mangle b=)
3. (mangle c=)
4. (mangle d=)
5. (mangle e=)
6. (mangle f=)
7. (mangle g=)
8. (mangle h=)

(mangle aob=)
(mangle cod=)
(mangle eof=)
(mangle aoh=)

1. (mangle rms=)
2. (mangle vmt=)
3. (mangle qmn=)
4. (mangle wpq=)
5. (mangle ajk=)
6. (mangle ckd=)
7. (mangle fkh=)
8. (mangle blc=)

Explanation:

Step1: Recall supplementary - angle property

Supplementary angles add up to 180°. For example, if one angle is \(x\) and its supplementary angle is \(y\), then \(x + y=180^{\circ}\).

Step2: Recall complementary - angle property

Complementary angles add up to 90°. If one angle is \(m\) and its complementary angle is \(n\), then \(m + n = 90^{\circ}\).

Step3: Solve for \(m\angle a\)

\(m\angle a\) and the 140° angle are supplementary. So \(m\angle a=180 - 140=40^{\circ}\).

Step4: Solve for \(m\angle b\)

\(m\angle b\) and \(m\angle a\) are vertical angles. Vertical angles are equal, so \(m\angle b = m\angle a=40^{\circ}\).

Step5: Solve for \(m\angle c\)

\(m\angle c\) and the 140° angle are vertical angles. So \(m\angle c = 140^{\circ}\).

Step6: Solve for \(m\angle d\)

\(m\angle d\) and the 119° angle are supplementary. So \(m\angle d=180 - 119 = 61^{\circ}\).

Step7: Solve for \(m\angle e\)

\(m\angle e\) and the 159° angle are supplementary. So \(m\angle e=180 - 159=21^{\circ}\).

Step8: Solve for \(m\angle f\)

\(m\angle f\) and \(m\angle e\) are vertical angles. So \(m\angle f = m\angle e=21^{\circ}\).

Step9: Solve for \(m\angle g\)

\(m\angle g\) and the 159° angle are vertical angles. So \(m\angle g = 159^{\circ}\).

Step10: Solve for \(m\angle h\)

\(m\angle h\) and \(m\angle d\) are vertical angles. So \(m\angle h = m\angle d=61^{\circ}\).

Step11: Solve for \(m\angle AOB\)

The sum of angles around a point is 360°. At point \(O\), \(m\angle AOB=360-(61 + 24+74 + 53)=360 - 212 = 148^{\circ}\).

Step12: Solve for \(m\angle COD\)

\(m\angle COD = 53^{\circ}\).

Step13: Solve for \(m\angle EOF\)

\(m\angle EOF=24^{\circ}\).

Step14: Solve for \(m\angle AOH\)

\(m\angle AOH = 61^{\circ}\).

Step15: Solve for \(m\angle RMS\)

\(m\angle RMS=90 - 21=69^{\circ}\).

Step16: Solve for \(m\angle VMT\)

\(m\angle VMT=180-(62 + 50)=180 - 112 = 68^{\circ}\).

Step17: Solve for \(m\angle QMN\)

\(m\angle QMN = 99^{\circ}\).

Step18: Solve for \(m\angle WPQ\)

\(m\angle WPQ=180-(62 + 50)=68^{\circ}\).

Step19: Solve for \(m\angle AJK\)

\(m\angle AJK=180-(49 + 45)=86^{\circ}\).

Step20: Solve for \(m\angle CKD\)

\(m\angle CKD=180-(49 + 55)=76^{\circ}\).

Step21: Solve for \(m\angle FKH\)

\(m\angle FKH=180-(47 + 55)=78^{\circ}\).

Step22: Solve for \(m\angle BLC\)

\(m\angle BLC=180-(45 + 47)=88^{\circ}\).

Answer:

1. \(m\angle a = 40^{\circ}\)
2. \(m\angle b = 40^{\circ}\)
3. \(m\angle c = 140^{\circ}\)
4. \(m\angle d = 61^{\circ}\)
5. \(m\angle e = 21^{\circ}\)
6. \(m\angle f = 21^{\circ}\)
7. \(m\angle g = 159^{\circ}\)
8. \(m\angle h = 61^{\circ}\)
9. \(m\angle AOB = 148^{\circ}\)
10. \(m\angle COD = 53^{\circ}\)
11. \(m\angle EOF = 24^{\circ}\)
12. \(m\angle AOH = 61^{\circ}\)
13. \(m\angle RMS = 69^{\circ}\)
14. \(m\angle VMT = 68^{\circ}\)
15. \(m\angle QMN = 99^{\circ}\)
16. \(m\angle WPQ = 68^{\circ}\)
17. \(m\angle AJK = 86^{\circ}\)
18. \(m\angle CKD = 76^{\circ}\)
19. \(m\angle FKH = 78^{\circ}\)
20. \(m\angle BLC = 88^{\circ}\)