Question : Two circles touch each other externally at points P and AB is a direct common tangent which touches the circles at A and B, respectively. $\angle APB$ is:
Option 1: 90°
Option 2: 45°
Option 3: 100°
Option 4: 80°
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Correct Answer: 90°
Solution :
$TA = TP$ (tangent from T)
$TB = TP$ (tangent from T)
Now in $\triangle ATP,TA = TP$
$\therefore \angle APT = \angle PAT$ (base angles of isosceles triangle)
And in $\triangle BTP,TB = TP$
$\therefore \angle BPT = \angle PBT $ (base angles of isosceles triangle)
Now, in $\triangle APB$,
$\angle APB + \angle PBA + \angle PAB = 180^\circ$ (angle sum property of triangle)
⇒ $\angle APB + \angle PBT + \angle PAT = 180^\circ$
⇒ $\angle APB + \angle BPT + \angle APT = 180^\circ\ (∵ \angle APT = \angle PAT$ and $\angle BPT = \angle PBT)$
⇒ $\angle APB + \angle APB = 180^\circ\ (∵\angle APB = \angle BPT + \angle APT)$
⇒ $2\angle APB = 180^\circ$
$\therefore\angle APB = 90^\circ$
Hence, the correct answer is 90°.
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