Question

# (i) In the given figure, lines AB and CD intersect at O, such that ∠AOD + ∠BOD + ∠BOC = 300°. Find ∠AOD. (ii) In the given figure, AB || CD, ∠APQ = 50° and ∠PRD = 120°. Find ∠QPR.

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Solution

## (i) We know that the sum of the angles around a point is $360°$. $\therefore \angle AOD+\angle BOD+\angle BOC+\angle AOC=360°\phantom{\rule{0ex}{0ex}}⇒\left(\angle AOD+\angle BOD+\angle BOC\right)+\angle AOC=360°\phantom{\rule{0ex}{0ex}}⇒300°+\angle AOC=360°\phantom{\rule{0ex}{0ex}}⇒\angle AOC=60°$ Also, $\angle AOC=\angle BOD=60°\left[\text{Vertically-Opposite Angles}\right]$ Now, $\angle AOD+\angle BOD=180°\left[\because AOB\text{is a straight line}\right]\phantom{\rule{0ex}{0ex}}⇒\angle AOD+60°=180°\phantom{\rule{0ex}{0ex}}⇒\angle AOD=120°$ (ii) $AB\parallel CD\text{and}PQ$ is the transversal. $\angle APQ=\angle PQR=50°\left[\text{Alternate Interior Angles}\right]$ Side QR of triangle PQR is produced to D. $\therefore \angle PRD=\angle QPR+\angle PQR\phantom{\rule{0ex}{0ex}}⇒120°=\angle QPR+50°\phantom{\rule{0ex}{0ex}}⇒\angle QPR=70°$

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