由外點引切線問題 急

1.
ta = tb (切線性質, tangent properties)
∠atb + ∠tba + ∠bat = 180° (三角形內角和, ∠s sum of △)
∠tba = ∠bat = x (等腰三角形底角, base ∠s, isos. △)
i.e. 70° + 2x = 180°
x = (180° - 70°)/2 = 55°

2.
ta = tb (切線性質, tangent properties)
x = 52° (等腰三角形底角, base ∠s, isos. △)
y
= 180° - 52° (直線上的鄰角, adj. ∠s on straight line)
= 128°

3.
ta = tb (切線性質, tangent properties)
∠atb + ∠tba + ∠bat = 180° (三角形內角和, ∠s sum of △)
∠tba = ∠bat = x (等腰三角形底角, base ∠s, isos. △)
i.e. 30° + 2x = 180°
x = (180° - 30°)/2 = 75°
ta // aO (切線⊥半徑, tangent ⊥ radius)
i.e. ∠taO = 90°
y = 90° - 75° = 15°

4.
因為 at // cb 和 ac // tb
所以 atbc 是一個平行四邊形(// gram)
則 ∠bat = ∠cab = x (平行四邊形對角線互相平分, diagonals of //gram)
又, at = tb (切線性質, tangent properties)
∠atb + ∠tba + ∠bat = 180° (三角形內角和, ∠s sum of △)
∠tba = ∠bat = x (等腰三角形底角, base ∠s, isos. △)
i.e. 60° + 2x = 180°
x = (180° - 60°)/2 = 60°

1.
at=bt (由外點引切線)
角tab=角tba=x (等腰三角形性質)
2x+70=180 (三角形內角和)
x=55

2.
ta=tb (由外點引切線)
x=52 (等腰三角形性質)
x+y=180 (直線上的鄰角)
52+y=180
y=128

3.
ta=tb (由外點引切線)
角tab=角tba=x (等腰三角形性質)
2x+30=180
2x=150
x=75
角oat=90 (切線垂直於半徑)
x+y=90
75+y=90
y=15

4.
ta=tb (由外點引切線)
at//bc (己知)
ac//bt (已知)
角abt=x (內錯角)
角tab=角tba (等腰三角形性質)
2x+60=180 (三角形內角和)
2x=120
x=60

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