In the given figure,$\mathrm{XY}=\mathrm{XZ} \mathrm{and} \mathrm{QY} \mathrm{isparallel} \mathrm{to} \mathrm{PM}.$
What is the value of $\mathrm{\angle }\mathrm{QRX}?$

Given, XY = XZ
$\therefore \mathrm{\angle }\mathrm{a}=\mathrm{\angle }\mathrm{b}$

Also,
$\begin{array}{l}\mathrm{\angle }\mathrm{a}+\mathrm{\angle }\mathrm{b}+\mathrm{\angle }\mathrm{YXZ}={180}^{\circ }\\ \Rightarrow 2\mathrm{\angle }\mathrm{b}+\mathrm{\angle }{40}^{\circ }={180}^{\circ }\\ \Rightarrow 2\mathrm{\angle }\mathrm{b}+{180}^{\circ }-{40}^{\circ }={140}^{\circ }\\ \Rightarrow \mathrm{\angle }\mathrm{b}={70}^{\circ }\end{array}$

Also,$\mathrm{QY}\parallel \mathrm{PM}$

$\begin{array}{r}\mathrm{\angle }\mathrm{QYM}+\mathrm{\angle }\mathrm{PMY}={180}^{\circ }\\ \Rightarrow \mathrm{\angle }\mathrm{QYM}={180}^{\circ }-{160}^{\circ }={20}^{\circ }\end{array}$

[interior angles on the same side of transversal are Supplementary]
$\begin{array}{l}\text{ In }\mathrm{△}\mathrm{RYZ},\\ \mathrm{\angle }\mathrm{XRY}\text{ is the exterior angle such that }\\ \mathrm{\angle }\mathrm{XRY}=\mathrm{\angle }\mathrm{RYZ}+\mathrm{\angle }\mathrm{RZY}={20}^{\circ }+{70}^{\circ }={90}^{\circ }\\ \text{ So, }\mathrm{\angle }\mathrm{QRX}+\mathrm{\angle }\mathrm{XRY}={180}^{\circ }\\ \Rightarrow \mathrm{\angle }\mathrm{QRX}+{90}^{\circ }={180}^{\circ }\\ \Rightarrow \mathrm{\angle }\mathrm{QRX}={90}^{\circ }\end{array}$