wongjack900

我第18 同 19題唔識做

18(a) ans:30度
19(b) ans:166.0度,346.0度

如看唔到去http://imageshack.us/photo/my-images/163/img001gd.jpg/

[ 本帖最後由 wongjack900 於 2011-7-26 10:17 PM 編輯 ]

rachel_luilui

18A  係AC+一條line segement 之後用angle in alt.segement  姐係angle abc=angle acq
and angle abc= angle aqc    let  angle abc be x     
since ab is diameter  angle bca = 90  
therefore  3x+90=180 (angles sum of triangle)
x=30

夏娜醬大好

開頭見到隻青蛙, 我仲以為係BIO功課添XD

18. (P係center?)
(a) <ABC = x
     <BAC = x
     <PBC = x
     2x = 90-x
     x = 30 (deg)
(b)(i) <QCD = 2x = 60 (deg)
          So, <CQD = <CDQ = <QCD = 60 (deg)
          So, CD = CD = BC
   (ii) <BQT = 180-30-60 = 90 (deg)
         So, ...
   (iii) <ACQ = x
         <BCA = (90+x)-x = 90 (deg)
         <ACD = 90 (deg)
         <ACD+<AQD = 180 (deg)
         So, ...
   (iv) Let M be the center of the lower circle.
         PC is penpendicular to RS.
         RS is penpendicular to CM.
         PCM is a stright line.
         <MCD = <PCB
         <MDC = <MCD = <PCB = <PBC
         BC = CD
         So, triangle PBC is congruent to triangle MDC.
         So, PC = MC.

19.
(a) (cosx+sinx)/(cosx-sinx)
      = (cosx+sinx)^2 / (cos^2 x - sin^2 x)
      = (1+2sinxcosx)/(1-2sin^2 x)
      = LHS
(b) (cosx+sinx)/(cosx-sinx) = 3/5
     2cosx+8sinx = 0
     tanx = -1/4