Question

The diagonals of a rectangle PQRS intersect at M. If $\angle$QMR = 50$°$ find $\angle$MPS.

Answer 1

Consider the given figure.

$$\begin{gathered} \mathrm{Given}: \\ \angle \mathrm{QMR}=50° \\ \angle \mathrm{PMS}=\angle \mathrm{QMR}=50°(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}) \\ \mathrm{Since},\mathrm{diagonals}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{are}\mathrm{equal}\mathrm{and}\mathrm{bisect}\mathrm{each}\mathrm{other} \\ \mathrm{As},\mathrm{PR}=\mathrm{SQ} \\ \frac{1}{2},\mathrm{PR}=\frac{1}{2}\mathrm{SQ} \\ \mathrm{Hence},\mathrm{MP}=\mathrm{MS} \\ \mathrm{So},\mathrm{in}∆\mathrm{PMS} \\ \angle \mathrm{MPS}=\angle \mathrm{MSP}=x\left(\mathrm{Angles}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}\right) \\ \mathrm{Now},\mathrm{consider}△\mathrm{PMS}. \\ \angle \mathrm{SPM}+\angle \mathrm{PMS}+\angle \mathrm{MSP}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow \mathit{}\mathrm{x}+50°+x=180° \\ \Rightarrow 2x+50°=180° \\ \mathit{\Rightarrow }x=\frac{130}{2} \\ \mathit{\Rightarrow }x=65° \\ \therefore \angle \mathrm{MPS}=\mathit{}\mathit{}x\mathit{}=65° \end{gathered}$$