e-maths qns..

• Fireindahouse

  4 Oct 07, 23:34

  
  
  can anyone kindly help me with this??
• eagle

  4 Oct 07, 23:46

  a) CDB = 18 degrees (isos. triangle)
  hence CFB = 18 degrees (angle at sector)
  
  b) CAB = 18 degrees (angle at sector)
  ACB = 90 degrees (angle at semicircle)
  hence ABC = 72 degrees
  
  c) DBE = 180 -18-18 = 144
  DBA = 180 - 144 = 36 degrees = 1/2 of CBA
  thus BD bisects
  
  d) CFB = 18 (part a)
  FCC = BDF = 51 (angle at sector)
  FBC = 180 - 51 - 18 = 111 degrees
  
  Pls help me check. Did mentally.
• seotiblizzard

  4 Oct 07, 23:47

  
• ^tamago^

  5 Oct 07, 11:26

  Originally posted by eagle:

  a) CDB = 18 degrees (isos. triangle)
  hence CFB = 18 degrees (angle subtended by same ard)
  
  b) CAB = 18 degrees (angle at sector)
  ACB = 90 degrees (angle at semicircle)
  hence ABC = 72 degrees
  
  c) DBE = 180 -18-18 = 144
  DBA = 180 - 144 = 36 degrees = 1/2 of CBA
  thus BD bisects
  
  d) CFB = 18 (part a)
  FCB = BDF = 51 (angle subtended by same ard)
  FBC = 180 - 51 - 18 = 111 degrees
  
  Pls help me check. Did mentally.

  eagle very smart. all correct, but one tiny error.
• tut4nkh4m3n

  5 Oct 07, 11:42

  Originally posted by ^tamago^:

  eagle very smart. all correct, but one tiny error.

  
• eagle

  5 Oct 07, 20:06

  Originally posted by ^tamago^:

  eagle very smart. all correct, but one tiny error.

  i already said I lousy in maths liao
  even so when everything do mentally on the comp
• Fireindahouse

  5 Oct 07, 22:59

  Originally posted by eagle:

  a) CDB = 18 degrees (isos. triangle)
  hence CFB = 18 degrees (angle at sector)
  
  b) CAB = 18 degrees (angle at sector)
  ACB = 90 degrees (angle at semicircle)
  hence ABC = 72 degrees
  
  c) DBE = 180 -18-18 = 144
  DBA = 180 - 144 = 36 degrees = 1/2 of CBA
  thus BD bisects
  
  d) CFB = 18 (part a)
  FCC = BDF = 51 (angle at sector)
  FBC = 180 - 51 - 18 = 111 degrees
  
  Pls help me check. Did mentally.

  ...
  
  thx !