SSC CPO SI and ASI 2012 Paper

Question : 133 of 200

Marks: +1, -0

O is the centre and arc ABC subtends an angle of 130° at O. AB is extended to P. Then PBC is

Solution:

(3)

\∠ A D C= {1}/{2} \× \∠ A O C

= {1}/{2} \× 130^{\°}=65^{°}

Since, ADCB is a cyclic quadrilateral and angle PBC is an exterior angle,
Therefore, ∠ PBC =∠ADC 65°

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