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Solution


Q.

Two circles touch each other externally at C and AB is common tangent of circles, then ∠ACB is
1 70°  
2 60°  
3 100°  
4 90°  

Ans.

(4)

Draw CM perpendicular to AB.

Now, AM = MC and MB = MC (tangents drawn from external point are equal).

⇒ AM = MC

⇒ ∠MAC = ∠MCA = 45°

(Since Δ AMC is right triangle)

∴  Also, MB = MC ⇒ ∠MBC = ∠MCB = 45° (Since Δ MBC is right angle triangle)

∴ ∠ACB = ∠MCA + ∠MCB = 45° + 45° = 90° ⇒ ∠ACB = 90°