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In Circle P, m EA=58, m BC=42, m CD=90, find:
1. m∠EDA
2. m∠BDC
3. m∠BEC
4. m arc AB
5. m∠AEB
6. m∠ECA
7. m∠BAC
8. m arc ED
9. m∠ECD
10. m∠ACB
11. m∠EBD
12. m∠ADB
13. m∠CAD
14. m∠EAB
15. m∠EBC
PLEASE HELP MU HUHU


Sagot :

Circle theorem states that the angle that an arc subtends at the center of a

circle is twice the angle subtended at the circumference.

The values are as follows;

  1. m∠EDA = 90°
  2. m∠BDC = 21°
  3. m∠BEC = 21°
  4. [tex]m\widehat{AB}[/tex] = 80°
  5. m∠AEB = 40°
  6. m∠ECA =  29°
  7. m∠BAC = 21°
  8. [tex]m\widehat{ED}[/tex]  =  90°
  9. m∠ECD = 45°
  10. m∠ACB = 40°
  11. m∠EBD = 45°
  12. m∠ADB = 40°
  13. m∠CAD = 45°
  14. m∠EAB = 90°
  15. m∠EBC = 90°

Reasons:

[tex]\displaystyle 1. \ \mathbf{m\angle EDA} = \frac{1}{2} \cdot m\angle EPC[/tex] Angle subtended at center is twice angle at circumference.

m∠EPC = 180°

∴  m∠EDA = 0.5 × 180° = 90°

2. [tex]m\widehat{BC}[/tex] = 42°, therefore, m∠BPC = 42°

[tex]\displaystyle \ m\angle BDC= \frac{1}{2} \cdot m\angle BPC[/tex]

m∠BDC = 0.5 × 42° = 21°

3. m∠BEC = m∠BDC = 21°

4. [tex]m\widehat{AB}[/tex] = 180° - (58° + 42°) = 80°

5. [tex]\mathbf{m\widehat{AB}}[/tex] = m∠APB by definition

m∠AEB = 0.5 × m∠APB

m∠AEB = 0.5 × 80° = 40°

6. m∠ECA = 0.5 × m∠AE

Therefore;

m∠ECA = 0.5 × [tex]m\widehat{EA}[/tex]

m∠ECA = 0.5 × 58° = 29°

m∠ECA =  29°

7. m∠BAC = 0.5 × [tex]\mathbf{m\widehat{BC}}[/tex]

m∠BAC = 0.5 × 42° = 21°

m∠BAC = 21°

8. [tex]m\widehat{ED}[/tex]  = 180° - [tex]\mathbf{m\widehat{CD}}[/tex]

[tex]m\widehat{ED}[/tex]  = 180° - 90° = 90°

[tex]\mathbf{m\widehat{ED}}[/tex]  =  90°

9. m∠ECD = 0.5 × [tex]\mathbf{m\widehat{ED}}[/tex]

m∠ECD = 0.5 × 90° = 45°

10. m∠ACB = 0.5 × [tex]m\widehat{AB}[/tex]

m∠ACB = 0.5 × 80° = 40°

11. m∠EBD = 0.5 × [tex]\mathbf{m\widehat{ED}}[/tex]

m∠EBD = 0.5 × 90° = 45°

12. m∠ADB = 0.5 × [tex]\mathbf{m\widehat{AB}}[/tex]

m∠ADB = 0.5 × 80° = 40°

13. m∠CAD = 0.5 × [tex]\mathbf{m\widehat{CD}}[/tex]

m∠CAD = 0.5 × 90° = 45°

14. m∠EAB = 0.5 ×[tex]\mathbf{m\widehat{EABC}}[/tex]

m∠EAB = 0.5 × 180° = 90°

15. m∠EBC = 0.5 ×[tex]\mathbf{m\widehat{EABC}}[/tex]

m∠EBC = 0.5 × 180° = 90°

Learn more about circle theorems here:

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