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Geometry

Cham

2007-07-06 2:10 am

In Fig 2.30, O1 and O2 are the centres of the smaller and the larger circles respectively. TA is a tangent to the smaller circle and SC is a common tangent to both circles. QCP and TABP are straight lines.
a) If angle QCA=72, find angle ACB.
b) Show that QH//KP.
c) If the radius of the larger circle is twice that of the smaller circle and QC=8cm, find AP. (Give the answer correct to 1 decimal place.)
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回答 (1)

[匿名]

2007-07-06 3:22 am

✔ 最佳答案

a)as TABP is the tangent of the smaller circle, thus angle QAT=angle QCA, also same, as SC is the tangent of both circles, let SC meet TABP to the point D, thus, angle ACD = angle AQC and angle DCB = angle CPB.

also we know that angela QAT=angle QCA=72, Thus, angle QAB=108 (180-72=108),

So angle AQC+ Angle CPB=180-108=72.

Thus angle ACB= angle ACD+ Angle DCB=angle AQC+ Angle CPB=72

b) sorry, i can't follow what u ask for this question

c)connect QO1, O1A, CO2, and O2P, thus, angela QO1A=2xangela QCA=2x72=144, and same as doing same thing in circle O2, now ,i think u know how to finish it.

karen

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