Q. 11 A24.2( 5 Votes )

Two circles touch each other externally at the point C. A direct common tangent AB touch the two circle at the points A and B. Value of ACB is
A. 60°

B. 45°

C. 30°

D. 90°

Answer :

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Let D be the point where the transverse tangent meets the direct tangent


DAC = DCA (Tangents drawn from an external point to the same circle are always equal and hence Δ DAC is isosceles)


Let DAC = DCA = a …Equation(i)


DBC = DCB (Tangents drawn from an external point to the same circle are always equal and hence Δ DBC is isosceles)


Let DBC = DBA = b …Equation(ii)


From Δ ABC we get


ACB = 1800-(a + b)


From Equation (i) and (ii) we get


DCA + DCB = (a + b)


ACB = (a + b)


Equating ACB found in the above two cases we get


1800-(a + b) = (a + b)


(a + b) = 900


So ACB = 900


So D is the correct option

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