help on this maths qns...

• Fireindahouse

  25 Oct 07, 22:25

  
  
  pls..
• eagle

  25 Oct 07, 23:25

  part a b c or d or all?
• Fireindahouse

  25 Oct 07, 23:58

  part a(4 and part c pls..
  
  answers provided are
  angle aeb=28degree
  
  part c. angle eba = 62degree.
• eagle

  26 Oct 07, 00:07

  Originally posted by Fireindahouse:

  part a(4 and part c pls..
  
  answers provided are
  angle aeb=28degree
  
  part c. angle eba = 62degree.

  part a) 4)
  
  angle ADB = 28 degrees (bisector)
  thus angle AEB = 28 degrees (angle at circumference)
  
  part c
  angle CBD = angle DAC (part ii u hv le) = 34 degrees
  angle XBD = 28 degrees (isos)
  
  angle CBA = from part iii ( I lazy to do )
  
  thus angle EBA = angle CBA - 34 - 28
• Fireindahouse

  26 Oct 07, 00:11

  Originally posted by eagle:

  part a) 4)
  
  angle ADB = 28 degrees (bisector)
  thus angle AEB = 28 degrees (angle at circumference)
  
  part c
  angle CBD = angle DAC (part ii u hv le) = 34 degrees
  angle XBD = 28 degrees (isos)
  
  angle CBA = from part iii ( I lazy to do )
  
  thus angle EBA = angle CBA - 34 - 28

  cannot assume like this...not equal bisector..
• eagle

  26 Oct 07, 00:18

  Originally posted by Fireindahouse:

  cannot assume like this...not equal bisector..

  DB bisects angle ADC. It is in your question
• popikachu

  26 Oct 07, 00:18

  a)
  i guess i and ii is very simple and don't need any explaination...
  
  a)i) "angle in semi circle"
  a)ii) "Total angle in Triangle"
  
  a)iii)
  Look at ABCD.
  All these point is in a circle so we can used 'opposite angles of cyclic quad'.
  180- ADC = ABC
  
  a)iv)
  Look at point A E X B D.
  Using 'angles in same segment', you can find AEB
  
  As DB bisect ADC, ADB is equal to 56/2 which is ADB
  
  ADB = AEB --> "angles in same segment"
  
  b)
  it is an isos triangle as there is 2 side with the same angle.
  XDB and XBD
  CDB=XBD
  
  c and d)
  XD and XB has the same length and this proves that both of them add up to be the diameter of a circle and each individual is a radius...
  This also proves x is the centre.
  
  EAB is 90 degree --> angle in semi circle
  180 - 90 - AEB = EBA
  
  Thats all... ^^
• ^tamago^

  26 Oct 07, 00:19

  ai) DCA = 90º (angle subtended by diameter)
  aii) DAC = 180º - 90º - 56º = 34º (angles in a triangle)
  aiii) CBA = DBA + DBC
  = 90º (angle subtended by diameter) + DAC (angle at circumference subtended by same arc)
  = 90º + 34º
  = 124º
  aiv) AEB = ADB (angle at circumference subtended by same arc)
  = ½ ADC
  = 28º
• Fireindahouse

  26 Oct 07, 00:26

  Originally posted by popikachu:
  a)
  i guess i and ii is very simple and don't need any explaination...
  
  a)i) "angle in semi circle"
  a)ii) "Total angle in Triangle"
  
  a)iii)
  Look at ABCD.
  All these point is in a circle so we can used 'opposite angles of cyclic quad'.
  180- A[b]DC = ABC
  
  a)iv)
  Look at point A E X B D.
  Using 'angles in same segment', you can find AEB
  
  As DB bisect ADC, ADB is equal to 56/2 which is ADB
  
  ADB = AEB --> "angles in same segment"
  
  b)
  it is an isos triangle as there is 2 side with the same angle.
  XDB and XBD
  CDB=XBD
  
  c and d)
  XD and XB has the same length and this proves that both of them add up to be the diameter of a circle and each individual is a radius...
  This also proves x is the centre.
  
  EAB is 90 degree --> angle in semi circle
  180 - 90 - AEB = EBA
  
  Thats all... ^^[/b]

  very confident!!
• absol

  26 Oct 07, 00:40

  answer all ba. but may nt all be correct...no guarantees^^
  
  a)i) angle DCA is 90, because it is an angle in a semicircle
  a)ii) angle DAC = 180-90-56= 24 deg, sum of angle in a triangle
  a(iii) angle CBA =180-56= 124 deg, opp angles in a cyclic quad are supplementary, (ABCD is a cyclic quad)
  a)iv) angle ADB = 28deg, therefore angle AEB = 28deg( angle in same seg)
  
  b) angle CDB = angle XBD(alt. angle)
  since angle BDX = angle XBD = 28deg, therefore triangle BDX is isoceles...
  
  c) angle DCA = angle DEA = angle DBA ( opposite angle in a cyclic quad are supplementary)( DEAB and EDCA are both cyclic quads)
  
  therefore angle DBA = 90deg
  angle EBA = 90-28=62deg
  
  d)this hence part..not very sure...
  
  edit: new way...but i think still wrong lol...not very much like "hence"
  
  angle EBA = 62deg
  angle DBA = angle in semicircle, 90deg
  therefore angle DBX = 90-62=28deg
  angle DXB = 124deg,
  so angle AXB = 56deg,
  since angle AEB and ADB are the angles on the circumference and subtended by the same arc as angle AXB and their angle is 28 with is half of AXB, therefore X mus be the middle of the circle as the rule angle at centre of circle is twice angle of circum is applied here...
  
  @popikachu
  quote"c and d)
  XD and XB has the same length and this proves that both of them add up to be the diameter of a circle and each individual is a radius...
  This also proves x is the centre.
  
  EAB is 90 degree --> angle in semi circle
  180 - 90 - AEB = EBA "
  
  i think ur method is wrong. u do d first den u do c. u are supposed to find part c angle first den hence,using the angle from c to find part d, not add the lengths up, prove, den do part c.
  plus i think u should state X is a point on the diameter too. so XB and XD will add up to the diameter. else i can plot a point anywhere on the circle and make an iso. triangle also.
• eagle

  26 Oct 07, 00:59

  dun think so cheem for part d
  
  EBA is 62 degrees
  AEB is 28 degrees
  
  Hence EAB is 90 degrees
  Due to angle at semicircle, EB is diameter
  
  EB (diameter) intersects with AD (also diameter) at X
  Hence, X is centre of circle
• absol

  26 Oct 07, 01:10

  i think ure right
• popikachu

  26 Oct 07, 12:35

  Originally posted by Fireindahouse:

  very confident!!

  
  
  So my answer correct mah?
• Fireindahouse

  26 Oct 07, 12:44

  Originally posted by popikachu:

  
  
  So my answer correct mah?

  ya
• felirox

  26 Oct 07, 14:53

  i hate the circles topic. it irks me.