# 12) in the figure, $\triangle P Q R$ is right at Q, $Q$, $11 \%$ and $\angle L m k=130^{\circ}$, $\angle L P M M L, R Q \angle P m h$ and $\angle D R Q$. \begin{aligned} \text { AOS } \angle L P R \bar{n} &=40^{\circ} \\ \angle P M L &=50^{\circ} \\ \angle P R Q &=50^{\circ} \end{aligned}

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12) in the figure, $\triangle P Q R$ is right at Q, $Q$, $11 \%$ and $\angle L m k=130^{\circ}$, $\angle L P M M L, R Q \angle P m h$ and $\angle D R Q$. \begin{aligned} \text { AOS } \angle L P R \bar{n} &=40^{\circ} \\ \angle P M L &=50^{\circ} \\ \angle P R Q &=50^{\circ} \end{aligned}

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(i) ∵ ML || RQ

and ∠LMR and ∠MRQ

are interior angles on the same side of transverse MR.

∴ ∠LMR + ∠MRQ = 180°

∴ ∠MRQ = 180° - ∠LMR

= 180° - 130°

= 50°

∴ ∠PRQ = ∠MRQ = 50°

(ii) ∠PML and ∠LMR are linear pair.

∴ ∠PML + ∠LMR = 180°

⇒ ∠PML = 180° - 130° = 50°

(iii) Now in triangle PQR, ∠PQR = 90°, ∠PRQ = 50°

And ∠QPR + ∠PQR + ∠PRQ = 180°

⇒ ∠QPR = 180° - ∠PQR - ∠PRQ

= 180° - 90° - 50°

= 180° - 140°

= 40°

Clearly ∠LPM = ∠QPR = 40°.