Question

# In Fig. 5, O is the centre of the circle. The angle by the arc BCD at the centre is 140°, BC is produced to P. Find ∠DCP.

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Solution

## We have, $\angle \mathrm{BOD}=140°$ Using the property of circles, the angle subtended by a chord at the center is twice the angle subtended by it on the remaining part of the circle, we get $\angle \mathrm{BAD}=\frac{1}{2}\angle \mathrm{BOD}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×140°\phantom{\rule{0ex}{0ex}}=70°$ Also, in cyclic quadrilateral ABCD, As, the opposite angles are supplementary. So, $\angle \mathrm{DCB}=180°-\angle \mathrm{BAD}\phantom{\rule{0ex}{0ex}}=180°-70°\phantom{\rule{0ex}{0ex}}=110°$ Now, $\angle \mathrm{DCP}+\angle \mathrm{DCB}=180°\left(\mathrm{Linear}\mathrm{pair}\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{DCP}=180°-\angle \mathrm{DCB}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{DCP}=180°-110°\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{DCP}=70°$

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