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.

Answer 1

(i)
We know that the sum of the angles around a point is $360°$.
$$\begin{gathered} \therefore \angle AOD+\angle BOD+\angle BOC+\angle AOC=360° \\ \Rightarrow \left(\angle AOD+\angle BOD+\angle BOC\right)+\angle AOC=360° \\ \Rightarrow 300°+\angle AOC=360° \\ \Rightarrow \angle AOC=60° \end{gathered}$$
Also,
$\angle AOC=\angle BOD=60°\left[\text{Vertically-Opposite Angles}\right]$
Now,
$$\begin{gathered} \angle AOD+\angle BOD=180°\left[\because AOB\text{is a straight line}\right] \\ \Rightarrow \angle AOD+60°=180° \\ \Rightarrow \angle AOD=120° \end{gathered}$$

(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.
$$\begin{gathered} \therefore \angle PRD=\angle QPR+\angle PQR \\ \Rightarrow 120°=\angle QPR+50° \\ \Rightarrow \angle QPR=70° \end{gathered}$$