Q.Two circles touch each other externally at point C. AB is a common tangent to both circles and touches the circles at points A and B, respectively. Find the measure of \(\angle\)ACB. (a) 60° (b) 45° (c) 30° (d) 90°
Answer: D
AP = PC
∴\(\angle\)PAC = \(\angle\)PCA

Similarly, \(\angle\)PBC = \(\angle\)PCB
\(\angle\)ACB = \(\angle\)PCB + \(\angle\)PCA = \(\angle\)PAC + \(\angle\)PBC

In ∆ABC,
\(\angle\)BAC + \(\angle\)ABC + \(\angle\)BCA = 180°
Or, \(\angle\)PAC + \(\angle\)PBC + \(\angle\)PAC + \(\angle\)PBC = 180°
Thus, \(\angle\)PAC + \(\angle\)PBC = 90°
∴\(\angle\)ACB = \(\angle\)PAC + \(\angle\)PBC = 90°
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