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Two equal circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB = 

A. 60° 

B. 45° 

C. 30° 

D. 90°

1 Answer

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Best answer

Answer is D. 90°

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. 

Property 2: Sum of all angles of a straight line = 180°

Property 3: Sum of all angles of a triangle = 180°. 

By property 1, ∆OAB is right-angled at ∠OAB (i.e., ∠OAB = 90°) and ∆PBA is right-angled at ∠ PBA (i.e., ∠ PBA = 90°) 

Clearly, 

∠b + ∠c = ∠OAB 

⇒ ∠b + ∠c = 90° 

⇒ ∠b = 90° - ∠c 

Similarly, 

∠d + ∠e = ∠PBA 

⇒ ∠d + ∠e = 90° 

⇒ ∠e = 90° - ∠d

Now, 

∠a = ∠b = 90° - ∠c [∵ OA = OC (Radius)] 

And, 

∠e = ∠f = 90° - ∠d [∵ PB = PC (Radius)] 

By property 2, 

∠a + ∠f + ∠ACB = 180° 

⇒ ∠ACB = 180° – ∠a – ∠f 

⇒ ∠ACB = 180° – (90° - ∠c) – (90° - ∠d) 

⇒ ∠ACB = 180° – 90° + ∠c – 90° + ∠d 

⇒ ∠ACB = ∠c + ∠d 

Now, in ∆ACB 

By property 3, 

∠ACB + ∠c + ∠d = 180° 

⇒ ∠ACB + ∠ACB = 180° [∵∠ACB = ∠c + ∠d] 

⇒ 2∠ACB = 180°

⇒ ∠ACB = \(\frac{180^°}{2}\)

⇒ ∠ACB = 90° 

Hence, ∠ACB = 90°

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