MATHS F.2 URGENT!!20點

Consider DAPQ:
sinθ = PQ/AP
sinθ = a/AP
AP = a/sinθ ...... (1)

PAD + PAQ = 90o
PAD + θ = 90o
PAD = 90o - θ

Consider DADP:
ADP + PAD + APD = 180o ( sum of D)
ADP + (90o - θ) + 90o = 180o
ADP = θ
sinADP = AP/AD
sinθ = AP/AD
AD = AP(1/sinθ)
But (1): AP = a/sinθ
Hence, AD = (a/sinθ)(1/sinθ)
AD = a/sin2θ ...... (2)

Consider DADC:
ACD + PAD + ADC = 180o ( sum of D)
ACD + (90o - θ) + 90o = 180o
ACD = θ
sinACD = AD/AC
sinθ = AD/AC
AC = AD(1/sinθ)
But (2): AD = a/sin2θ
Hence, AC = (a/sin2θ)(1/sinθ)
AC = a/sin3θ ...... (3)

Consider DADC:
cosACD = DC/AC
cosθ = DC/AC
DC = ACcosθ
But (3): AC = a/sin3θ
Hence, DC = (a/sin3θ)cosθ
DC = acosθ/sin3θ ...... (4)

Consider DDPC:
cosACD = PC/DC
cosθ = PC/DC
PC = DCcosθ
But (4): DC = acosθ/sin3θ
Hence, PC = (acosθ/sin3θ)cosθ
PC = acos2θ/sin3θ

Ans: AC = a/sin3θ and PC = acos2θ/sin3θ
=

Note that ㄥPAQ = ㄥADP = ㄥACD = θ
AP = a / sinθ
AP = a / sinθ / AD = sinθ , AD = a / sin^2θ
AD = a / sin^2θ / AC = sinθ ,
AC = a / sin^3θ
PC = AC - AP = a / sin^3θ - a / sinθ
= (a - a sin^2θ ) / sin^3θ
= a( 1 - sin^2θ ) / sin^3θ
= a cos^2θ / sin^3θ