Maths(Tangent to circles)題目

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,


圖片參考：http://i178.photobucket.com/albums/w276/calligraphy-yahoo2006/ScreenHunter_propertiesofcircle.jpg

∠PAT + ∠QBD = 90° ( tangent ⊥ radius )
∠QBD = 10°
∠APB = 90° ( ∠ in semi circle )
∠BPQ = ∠QBD ( ∠s in the same segment)
∠BPQ = 10°
∠APT + ∠APB + ∠BPQ = 180° ( adj. ∠s on st.line )
∠APT + 100° = 180°
∠APT = 80°
∠ATP = ∠TAP ( base ∠s, isos. △)
In △ATP,
∠APT + ∠ATP + ∠TAP = 180° ( ∠ sum of △)
80° + 2( ∠ATP ) = 180°
2( ∠ATP ) = 100°
∠ATP = 50°
In △ATQ,
∠TAQ + ∠ATQ + ∠TQA = 180° ( ∠ sum of △)
∠TQA + 80° + 50° = 180°
∠TQA = 50°
∵ ∠TQA = ∠ATQ
∴ AT = AQ ( sides opp. eq. ∠s )
(b) find
(i) ∠ATP,
In △ATP,
∠APT + ∠ATP + ∠TAP = 180° ( ∠ sum of △)
80° + 2( ∠ATP ) = 180°
2( ∠ATP ) = 100°
∠ATP = 50°
(ii) ∠OPB
∠PBA = ∠PQA ( ∠s in the same segment)
∠PBA = 50°
OP = OB ( radii of circle )
∠PBA = ∠OPB ( base ∠s, isos. △)
∠OPB = 50°

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