(附圖在http://aerodrive.twghwfns.edu.hk/~4s227/maths_ttc_2.bmp)
In the figure, AB is a diameter of the semi-circle. TS is the tangent to the semi-circle at A. Given that TPQ is a straight line, ∠TAQ = 80° and PT = PA,
(a) prove that AT = AQ,
(b) find
(i) ∠ATP,
(ii) ∠OPB.
Maths(Tangent to circles)題目
2007-03-11 4:26 am
回答 (1)
2007-03-11 5:17 am
✔ 最佳答案
a)∠PAT=∠PBA(∠s in alt. segment)∠PBA=∠PQA(∠s in the same segment)
∠PAT=∠PTA(base ∠s, isos. triangle)
so ∠ATQ=∠AQT,
AT=AQ(sides opp. eq. ∠s).
bi)by a, ∠ATP=(180-80)/2(∠ sum of triangle)
=50
bii)by bi, ∠PAT=∠ATP=50(base ∠s, isos. triangle)
∠PBO=∠PAT=50(∠s in alt segment)
PO=PB(radii)
so, ∠OPB=∠PBO=50(base∠s, isos. triangle)
∠OPB=50
收錄日期: 2021-04-16 16:47:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070310000051KK04492