19. In the figure, A is a point on the circumstance. OB intersects the circumstance at C and AC = OC= CB.
a) find angle CAB.
b) Prove that AB is the tangent to the circle at A.
圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701010100127313873376710.jpg
MATH QUESTION CIRCLES
2010-10-11 2:40 am
回答 (2)
2010-10-11 2:56 am
✔ 最佳答案
(a) angle CAB = angle CBA = xangle OCA=angle CAB + angle CBA =2x
Since triangle OAC is an equilateral triangle
angle COA = angle OAC = CBA = 2x
3(2x)=180=>x=30
(b) angle OAB=angle OAC + angle CAB = 90
So AB is the tangent to the circle at A.
2010-10-11 3:01 am
a)
OC = AC (Given)
OC = OA (radius)
So
OC = AC = OA
△OAC is a equilateral △.
So ㄥOCA = 60°
AC = BC (Given)
So ㄥCAB = ㄥCBA (base ∠s, isos. Δ)
We have
ㄥCAB + ㄥCBA = ㄥOCA (ext. ∠ of Δ)
2ㄥCAB = 60°
ㄥCAB = 30°
b)
ㄥOAB = ㄥOAC + ㄥCAB
ㄥOAB = 60° + 30° = 90°
AB is the tangent to the circle at A.(tangent⊥radius)
OC = AC (Given)
OC = OA (radius)
So
OC = AC = OA
△OAC is a equilateral △.
So ㄥOCA = 60°
AC = BC (Given)
So ㄥCAB = ㄥCBA (base ∠s, isos. Δ)
We have
ㄥCAB + ㄥCBA = ㄥOCA (ext. ∠ of Δ)
2ㄥCAB = 60°
ㄥCAB = 30°
b)
ㄥOAB = ㄥOAC + ㄥCAB
ㄥOAB = 60° + 30° = 90°
AB is the tangent to the circle at A.(tangent⊥radius)
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