圓的基本性質問題 急

adj. angles on straight line = 直線上的鄰角
angles sum of triangle = 三角形內角和
base angles, isos. triangle = 等腰三角形底角
Pyth. thm. = 畢氏定理
rejected = 捨去
since = 因為
tangent ⊥ radius = 切線⊥半徑
tangent properties = 切線性質

1.
TA = TB (tangent properties)
∠A = ∠B (base angles, isos. triangle)
∠T + ∠A + ∠B = 180° (angles sum of triangle)
x + 2∠B = 180°
∠B = (180° - x)/2
->
(180° - x)/2 + 5x = 180° (adj. angles on straight line)
180° - x + 10x = 360°
9x = 180°
x = 20°

2.
Since OB = radius = OA = x
OT = x + (x + 1) = 2x + 1\
Since OA⊥AT (tangent ⊥ radius)
OA^2 + AT^2 = OT^2 (Pyth. thm.)
x^2 + 15^2 = (2x + 1)^2
x^2 + 225 = 4x^2 + 4x + 1
3x^2 + 4x = 224
x^2 + 4x/3 = 224/3
(x + 2/3)^2 - 4/9 = 224/3
(x + 2/3)^2 = 676/9
x + 2/3 = +/- 26/3
x = 24/3 = 8 or x = -28/3 (rejected)
x = 8

3.
Since OA⊥PQ (tangent ⊥ radius)
OA^2 + AQ^2 = OQ^2 (Pyth. thm.)
(c/2)^2 + 15^2 = 17^2
(c/2)^2 = 64
c/2 = 8 (since c > 0)
c = 16

4.
Since OC = radius = OA = 2x
OT = 3x

Since OA⊥AT (tangent ⊥ radius)
OA^2 + AT^2 = OT^2 (Pyth. thm.)
(2x)^2 + 125 = (3x)^2
125 = 5x^2
x^2 = 25
x = 5 (since x > 0)

1.
TA=TB (由外點引切線)
角TAB=角TBA (等腰三角形性質)
角TBA=180-5x (直線上的鄰角)
x+2(180-5x)=180 (三角形內角和)
x+360-10x=180
-9x=-180
x=20

2.
OA垂直TA (切線垂直半徑)
OA=OB (半徑)
x^2+15^2=(x+1+x)^2 (畢氏定理)
x^2+225=(2x+1)^2
x^2+225=4x^2+4x+1
0=3x^2+4x-224
x=8 或 x=-28/3 (括去)

3.
OA垂直PQ (切線垂直半徑)
OA^2+AQ^2=OQ^2 (畢氏定理)
OA^2+15^2=17^2
OA=8 cm
OA=OB=8 (半徑)
所以 AB=8+8=16cm

4.
OA垂直TA (切線垂直半徑)
OC=OA (半徑)
TA^2+AO^2=TO^2 (畢氏定理)
125+2(x)^2=(3x)^2
125=9x^2-4x^5
125=5x^2
x=5