maths F.4 20點 new

Q29. For triangle ATB and triangle CTA.
Angle ATB = angle ATC ( common)
angle TAB = angle ACT ( angle in alt. segment)
therefore triangle ATB is similar to triangle CTA (AAA).
Since the 2 triangles are similar, TA/TC = TB/TA, so 12/15 = TB/12
TB = 144/15 = 9.6, so BC = TC - TB = 15 - 9.6 = 5.4
Q31.
Angle BQC = angle BPC = 90 degree. Therefore, BCPQ are concyclic because angles in the same segment are equal.
Angle QBC = angle QPA ( ext. angle of cyclic quad.)
But angle QPA = angle TAP ( alt. angle, TA//QP)
therefore, angle QBC = angle TAP
therefore TA is a tangent to the circle because angles in alt. segment are equal.

2010-04-08 12:00:41 補充：
Q25. Ag QBO = Ag QDO ( base Ag of isos. tri BOD). In tri DRO, Ag ODR = Ag QDR - Ag QDO = 45 - Ag QDO. In tri OPB, Ag POB + Ag QBO = 45 ( ext. Ag of tri), so Ag POB = 45 - Ag QBO. So Ag POB = Ag ODR. So tri OPB = tri DRO (AAS). PB = OR = sqrt [ OD^2 - DR^2] = 6. Note: Ag means angle.

2010-04-08 12:07:23 補充：
Q25. Since angle PQB = 45, therefore angle PBQ = 90 - 45 = 45. Triangle is therefore an isos. triangle, so PQ = PB = 6. OP = DR = 8, so OQ = OP - PQ = 8 - 6 = 2.

2010-04-08 12:22:03 補充：
Q23. Angle ABC = angle ACB ( base angle of isos. triangle ABC). Angle ABC = angle ADC ( angle in same segment), so angle ACB = angle ADC. Angle EAC = angle DAC ( common), so triangle ACE similar to triangle ADC (AAA). So AC/AE = AD/AC, 18/16 = AD/18, AD = 20.25, so ED = AD - AE = 20.25 - 16 = 4.25.

2010-04-08 12:32:45 補充：
Q21. AOOC is the diameter, so angle ABC = angle ODC = 90 ( angle in semi-circle). So AB//OD. Since AO = OC, so DC = BD = 4 ( mid - point theorem). AB = sqrt [ AC^2 - BC^2] = sqrt(16^2 - 8^2) = sqrt 192. = 8 sqrt 3. Angle DCA = arctan (AB/BC) = arctan sqrt 3 = 60 degree.