Discussion

Explanation:

In ∆PAB,

PA = PB    …(tangents from an external point to a circle are congruent)

∴ m ∠PBA = m ∠PAB; …(angles opposite to congruent sides)

∴ m ∠PBA = 60° …(sum of angles in a triangle is 180°)

Similarly, in ∆QBC 

m ∠CBQ = m ∠BCQ = 50°

∵ m ∠PBA + m ∠CBQ + x = 180° …(linear angles)

∴ x = 180° – 60° – 50° = 70°

Now, m ∠AOC = 2 × x

∴ m ∠AOC = 140°

Hence, option (e).

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