F.4 more about circles

[匿名]

2010-05-21 2:40 am

1)
http://i729.photobucket.com/albums/ww291/kaka1307/DSC00536.jpg
AT is the tangent to the circle at point T. OBA is a straight line and
angle ATB =x
a)express angle TAB in terms of x.
b) If angle TAB=3 angle ATB.find angle ATB
ans: 1a)90-2x 1b)18

2)
http://i729.photobucket.com/albums/ww291/kaka1307/DSC00537.jpg
PT = QT, S is the mid-point of PQ, ST is a diameter of the circle.
Determine whether PQ touches the circle at ponit S.

3)
http://i729.photobucket.com/albums/ww291/kaka1307/DSC00536.jpg
AB is the tangent to the circle at point C. Given that angle BOA=90,
OB=12 and OA=16,find the length of OC
the ans is 9.6

pls show clear steps.thz

回答 (1)

STEVIE-G™

2010-05-21 5:39 am

✔ 最佳答案

1a)
angle OTA=90 (radius is perpendicular to tangent)

In tri. OBT,
angle OTB=angle OBT=90-x
angle TOB=180-2(90-x)=2x

In tri. OAT,
angle TAB=180-90-angle TOB=90-2x

b)
angle TAB=3 angle ATB
90-2x=3x
5x=90
x=18

2)
In tri. TPS and tri. TQS
PT=QT
ST=ST (common side)
PS=QS

tri. TPS=tri. TQS (S.S.S.)

angle TSP=angle TSQ=90
so, PQ is a tangent to the circle
PQ touches the circle at S

3)
The figure is for Q1...

2010-05-20 22:28:43 補充：
3)
In tri. OAB,
OC=√(12^2+16^2)=20

angle OCA=angle OCB=90

In tri. OAC,
OC^2+AC^2=16^2=256...(1)

In tri. OCB,
OC^2+CB^2=144
OC^2+(20-AC)^2=144...(2)

(1)-(2):
AC^2-(20-AC)^2=256-144
40AC-400=112
AC=12.8

Sub AC=12.8 into (1)
OC^2+12.8^2=256
OC=9.6cm

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