amaths!!!!!!!!!

(a)
AP/sinθ=CP/sinA,PB/sin2θ=CP/sinB
∵A=B,PB=AB-AP=1-AP
∴AP/sinθ=(1-AP)/sin2θ
AP/sinθ=(1-AP)/2sinθcosθ
AP=(1-AP)/2cosθ
2cosθ*AP=1-AP
2cosθ*AP+AP=1
AP(1+2cosθ)=1
AP=1/(1+2cosθ)

(b)
∵AP=1/(1+2cosθ)
C=3θ
0 < 3θ < 180°
0 < θ < 60°
cos0=1
cos60°=1/2
1/2 < cosθ < 1
1+2*1/2 < 1+2cosθ < 1+2*1
2 < 1+2cosθ < 3
1/2 > 1/(1+2cosθ) >1/3
∴1/3 < AP < 1/2

(a) Use the sine formula to prove that AP = 1 / (1+2cos θ )
In triangle ACP,
AP/sinθ = CP / sin angle CAP
CP = AP sin angle CAP / sin θ……..1

In triangle BCP,
CP / sin angle CBP = BP / sin 2θ
BP = 1 – AP ***
1-AP = CP sin 2θ / sin angle CBP
CP = (1-AP) sin angle CBP / sin 2θ………..2

from 1 and 2,
AP sin angle CAP / sin θ = (1-AP) sin angle CBP / sin 2θ
angle CAP = angle CBP ( isosceles triangle, AC = BC ) ***
AP / sinθ = (1-AP) / sin 2θ
AP sin 2θ = (1-AP) sinθ
AP x 2 sinθ cosθ = sinθ – AP sinθ
AP x 2cosθ = 1 – AP
AP ( 1+2cosθ) = 1
AP = 1/(1+2cosθ) (shown)

(b) Asθvaries , using the result in (a) , show that 1/3 < AP < 1/2
AP = 1/(1+2cosθ)
-1≦cosθ≦1
-2≦2cosθ≦2
-1≦1+2cosθ≦3
1/(-1)≧1/(1+2cosθ)≧1/3
-1≦AP≦1/3

2007-03-23 23:41:17 補充：
angle ACB = θ 2θ = 3θ 0° 1/(1 2cosθ) 1/31/3