Two circles touch each other externally at C. AB is a direct common tangent to the two circles, A and B are points of contact; and ∠CAB = 55°. Then ∠ACB is:
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 90°.
Given that
Two circles touch each other externally at C
AB is a direct common tangent, ∠CAB = 55°
Draw a another common tangent through a common point C
PA = PC, PB = PC
∠PAC = ∠PCA = 55°
In triangle PBC
Let ∠PBC = ∠PCB = α°
∠ACB = ∠PCA + ∠PCB = α° + 55°
In triangle ABC
∠ACB + ∠BAC + ∠ABC = 180∘
⇒ α° + 55° + 55°+ α° = 180∘
⇒ 2α° = 180∘ - 110° = 70°
⇒ α° = 35°
Hence, ∠ACB = 55° + 35° = 90°