In the given figure, PBQ is a straight line while PA, PB, QB, and QC are tangents to the circle with center O. Find the value of angle AOC.
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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