Question : 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 $\angle CAB = 55^{\circ}$. Then $\angle ACB$ is:
Option 1: $35^{\circ}$
Option 2: $90^{\circ}$
Option 3: $55^{\circ}$
Option 4: $45^{\circ}$
Correct Answer: $90^{\circ}$
Solution :
Given: Two circles touch each other externally at C, and $\angle CAB = 55^{\circ}$.
Tangents drawn from an exterior point to a circle have equal lengths.
PA = PC
⇒ $\angle PCA= \angle PAC=x$
Also, PC = PB.
⇒ $\angle PBC= \angle PCB=y$
In $\triangle ABC$, $\angle ABC+ \angle ACB+ \angle BAC=180^{\circ}$.
⇒ $y+x+y+x=180^{\circ}$
⇒ $2(y+x)=180^{\circ}$
⇒ $y+x=90^{\circ}$
⇒ $\angle ACB=90^{\circ}$
Hence, the correct answer is $90^{\circ}$.
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