數學知識交流 － 三角 (難)

∠PCQ = 30°

2011-12-31 11:25:50 補充：
Let ∠CAQ be 2θ , then :

ㄥQPC = ∠CAQ + ㄥPQA = 2θ + 2θ = 4θ
and
ㄥQPB + ㄥQBP = ㄥPQA = 2θ
then ㄥQPB = ㄥQBP = θ

So
ㄥCPB = ㄥQPC - ㄥQBP = 4θ - θ = 3θ

On the other hand ,
ㄥCBA = (180° - ∠CAQ ) / 2 = (180°- 2θ) / 2 = 90°- θ
So
ㄥCBP = ㄥCBA - ㄥQBP = (90°- θ) - θ = 90°- 2θ

Let AP = PQ = QB = BC = k ,
then
(AQ/2) / AP = cos 2θ
So
AQ = 2k cos 2θ = PC

Consider △PBC , by sine formula :

sinㄥCPB / BC = sinㄥCBP / PC
sin 3θ / k = sin(90°- 2θ) / (2k cos 2θ)
sin 3θ = cos 2θ / (2cos 2θ)
sin 3θ = 1 / 2
3θ = 30° since θ < 90°
θ = 10°

Therefore
∠CAQ = 2θ = 20°
ㄥCBQ = (180 - 20)/2 = 80°
ㄥBCQ = (180 - 80)/2 = 50°

∴ ㄥPCQ = 180 - 20 - 80 - 50 = 30°

Hints :

Let ∠CAQ be 2a , then
... ...

If angle CBQ = b,
then angle CAQ = 180-2b
and angle CQB = 90 - b/2
By angle CQB = angle QAC + angle QCA
then angle QCA = 90 - b/2 - 180 + 2b = 3b/2 - 90???