Maths(Tangent to circles)題目

[匿名]

2007-03-11 4:00 am

(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.

回答 (1)

用戶82640

2007-03-11 5:07 am

✔ 最佳答案

a)
∠TAP=∠ATP(base∠,isos triangle)
∠PQA=∠TAP(∠ in alt. segment)
so,∠ATP=∠PAQ
also,AT=AQ(sides opp. eq.∠)

b)(i)
∠TAP+∠ATP+∠PQA=180(∠ sum of triangle)
80+∠ATP+∠PQA=180
2∠ATP=100 (by a,∠ATP=∠PAQ)
∠ATP=50

(ii)
∠TAQ+∠QAB=90 (tangent perpendicular to radius)
80+∠QAB=90
∠QAB=10

∠PTA=∠TAP(base∠,isos triangle)
∠TAP=50

∠TAP+∠PAQ=80
50+∠PAQ=80
∠PAQ=30

AO=PO(radius)
∠PAO=∠APO(base∠,isos triangle)
∠PAQ+∠QAB=∠APO
30+10=∠APO
∠APO=40

∠APB=90(∠ in semicircle)
∠APO+∠OPB=90
40+∠OPB=90
∠OPB=50

p.s.因為想你明白d,所以將所有步驟打晒出黎,希望唔好嫌麻煩
同埋未必岩晒 =.=

參考: 自己

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