高二三角函數數學題?

圓的兩切線交點 P (正方形左下頂點) 與圓
心 O 連線, 與兩切線之夾角設為 θ. 則
PO = √(2o^2+1oo^2)
sin(θ) = 2o/PO, cos(θ) = 1oo/PO.

兩切線間夾角為 2θ.
sin(2θ) = 2sin(θ)cos(θ) = 5/13

設 AB線段與正方形底邊相文於 R. 又令正
方方形右下頂點是 C. 則
    △ARC 相似 △PRB.
∴ AR : PR = RC : RB = AC : PB

設 RC = x, 則
AR = √(60^2+x^2),
PR = 60 - x ,
RB = PR × sin(2θ) = 5(60-x)/13

∴ √(60^2+x^2) : 60-x = x : 5(60-x)/13
∴ √(60^2+x^2) = x/(5/13) = (13/5)x
∴ 60^2 + x^2 = (169/25)x^2
∴ x = 25

AB = AR + RB
      = 65 + 5(35)/13 = 1020/13(cm)