Question

Two circle touch each other externally at C and AB is a common tangent to the circles. Then, $\angle$ACB = ______.

• A. 60°
• B. 45°
• C. 30°
• D. 90°

Answer 1

The correct option is D

90°

We know that $\angle$OAB = $\angle$PBA = ${90}^{\circ }$

Also $\angle$OAC = $\angle$OCA and $\angle$ PBC = $\angle$PCB

$\angle$AOC = ${180}^{\circ }$ – 2 $\angle$OCA ($\angle$AOC + $\angle$OAC + $\angle$OCA = ${180}^{\circ }$)

Similarly $\angle$BPC = ${180}^{\circ }$ – 2$\angle$BCP

Now $\angle$AOC + $\angle$BPC + $\angle$OAB + $\angle$PBA = ${360}^{\circ }$

${180}^{\circ }$ – 2 $\angle$OCA + ${180}^{\circ }$ – 2$\angle$BCP + ${90}^{\circ }$ + ${90}^{\circ }$ = ${360}^{\circ }$

Therefore $\angle$OCA + $\angle$BCP = ${90}^{\circ }$

For the straight line OCP, $\angle$OCA + $\angle$BCP + $\angle$ACB = ${180}^{\circ }$

${90}^{\circ }$ + $\angle$ACB = ${180}^{\circ }$

Therefore $\angle$ACB = ${90}^{\circ }$