Geometry!

• secretliker

  13 Apr 08, 00:13

  

  Tell me how to solve and which rules of geometry to apply. It's sec 4 maths, thanks!

• FirePig

  13 Apr 08, 00:22

  Use opp angles of cyclic quad, use adjacent angles on straight line, cosine rule maybe, sine rule

   

  find area of sector (1/2r^2theta*0.7/2pi) - area of triangle Area(a*b*sin(C)/2)...

   

  I will leave the details another day. You go try first. 

• eagle

  13 Apr 08, 00:43

  a)
  angle abc = pi - 1.38 rad

  using sine rule,

  AB/sin0.7 = 10/sin(pi-1.38)
  AB = 6.56cm

   

  b) angle oab = pi - (0.7) - (pi-1.38) = 0.68 rad
  area of shaded area = area of sector - area of unshaded triangle

  = 1/2 (10^2) (0.7) - 1/2 (10)(6.56)sin(0.68)
  = 35 - 20.6244
  =14.4 cm^2

• secretliker

  13 Apr 08, 00:57

  Thanks eagle. but part b first line should be:

  angle OAB = pi - angle AOB - angle ABO = pi - 0.7 - 1.38 = 1.06159 rad

• weiwei

  15 Apr 08, 10:58

  Originally posted by eagle:

  a)
  angle abc = pi - 1.38 rad

  using sine rule,

  AB/sin0.7 = 10/sin(pi-1.38)
  AB = 6.56cm

   

  b) angle oab = pi - (0.7) - (pi-1.38) = 0.68 rad
  area of shaded area = area of sector - area of unshaded triangle

  = 1/2 (10^2) (0.7) - 1/2 (10)(6.56)sin(0.68)
  = 35 - 20.6244
  =14.4 cm^2

  eagle, think there's a minor typo in part (a) leh.

  using sine rule,

  AB/sin0.7 = 10/sin1.38

  your answer correct >.<

• eagle

  15 Apr 08, 11:05

  Originally posted by weiwei:

  eagle, think there's a minor typo in part (a) leh.

  using sine rule,

  AB/sin0.7 = 10/sin1.38

  your answer correct >.<

  yup TS has already pointed out... He's a good self-learner...

  the answer appears correct because sin(1.38) gives the same value as sin(pi - 1.38)