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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