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ANGLE AND LINE

ANGLE AND LINE. By: NIKMATUL HUSNA. SUPPLEMENTARY. AOC +  BOC =  AOB a 0 + b 0 = 180 0 We can write a 0 = 180 0 - b 0 or b 0 = 180 0 - a 0. Conclusion :. The sum of supplementary angle is 180 0. An angle is the supplement of another angle. COMPLEMENTARY.

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ANGLE AND LINE

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  1. ANGLE AND LINE By: NIKMATUL HUSNA

  2. SUPPLEMENTARY AOC +  BOC =  AOB a0 + b0 = 1800 We can write a0 = 1800 - b0 or b0 = 1800 - a0

  3. Conclusion : • The sum of supplementary angle is 1800. • An angle is the supplement of another angle

  4. COMPLEMENTARY PQS +  RQS =  PQR X0 + Y0 = 900 We can write X0 = 900 - Y0 or Y0 = 900 - X0

  5. Conclusion : • The sum of complementary angle is 900. • An angle is the complement of another angle

  6. OPPOSITE ANGLE • POQ =  SOR • POS =  QOR • POQ and  SOR are opposite angle. • POS and  QOR are opposite angle.

  7. Conclusion : • The measure of opposite angle is same

  8. Opposite Angles 1and 7 2and 8 3and 5 4and 6 5 6 4 7 8 3 2 t 1 1 = 7, 2= 8, 3= 5, 4= 6

  9. 5 6 4 7 8 3 2 t 1 • Corresponding Angles • Alternate Interior Angles • Alternate Exterior Angles • Consecutive Interior Angles • Consecutive Exterior Angles

  10. Corresponding Angles 4 and 2 3 and 1 5 and 7 6 and 8 5 6 4 7 8 3 2 t 1 4 = 2, 3 = 1, 5 = 7, 6 = 8

  11. Alternate Interior Angles 3 and 7 2 and 6 5 6 4 7 8 3 2 t 1 3 = 7 and2 = 6

  12. Alternate Exterior Angles 5 and 1 4 and 8 5 6 4 7 8 3 2 t 1 5 = 1 and4 = 8

  13. Consecutive Interior Angles 5 3 and 2 6 and 7 6 4 7 8 3 2 t 1 3 +2 = 1800 and 6 +7 = 1800

  14. Consecutive Exterior Angles 5 4 and 1 5 and 8 6 4 7 8 3 2 t 1 4 + 1 = 1800 and 5 + 8 = 1800

  15. 5 6 4 7 8 3 2 t 1 Example : Look at the picture. Given 1 = 750 and Find: • 1 5 • 2 6 • 3 7 • 4 8

  16. Answer : • 1 = 7505 = 750 • 2 =1050 6 = 1050 • 3=750 7= 750 • 4=1050 8= 1050

  17. 5 6 4 7 8 3 2 t 1 Example : Look at the picture. Given 1 = (3x+45)0 and 4 =(5x+23)0 . Find: • 1 5 • 2 6 • 3 7 • 4 8

  18. Answer : • 1 +4 =1800 (3x+45)0 + (5x+23)0 =1800 3x0+ 5x 0 +45 0 + 23 0 =1800 8x0 =1800- 450 – 230 8x0 =112 0 x = 140

  19. 1=(3x+45)0= 3 X 140 +450 = 420 +450 = 870  4=1800- 870 = 930  2= 930  3= 870  5= 870  6= 930  7= 870  8= 930

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