2emaths qns..

• Fireindahouse

  4 Oct 07, 23:47

  
  
  part c..
  
  &
  
  Qusn2.
  
  many thks for any help pls//
• eagle

  5 Oct 07, 00:00

  a) CBD = BDA (alternate angles)
  ACB = BDA (angle at sector)
  Hence BCP = CBD, BPC is isoseles
  
  b) BDA = ACB = 32
  ABD = 180 - 118 - 32 = 30
  ACD = ABD (angle at sector) = 30 degrees
  
  c)
  i)
  DCF = 180 - 110 = 70 (opp angles in cyclic quad)
  BCD = ACD + BCA = 30 + 32 = 62
  CDF = 62 degrees (alternate angles)
  CFD = 180 - 62 - 70 = 48
  CED = CFD (angle at sector) = 48
  CBD = 32
  BCE = 180 - 32 - 48 = 100
  ACE = BCE - ACB = 68 degrees
  
  Someone check this... I think my method very inefficient
  
  ii)
  COD = 1/2 CED = 24 degrees
• eagle

  5 Oct 07, 00:04

  
  
  x = 180 - 69 = 111 degrees (opp angles in cyclic quad)
• HyuugaNeji

  5 Oct 07, 01:04

  cbd=32 (by isos triange)
  fde=32 (parallel lines) (need to assume is one straight line...otherwise your method more correct)
  fce=32
  dce=dcf-fce
  =70-32
  38
  
  ace=acd+dce
  =30+38
  =68