In the rectnagle PQRS shown below if angle PSM = 70 degree then find A

1.	In the rectnagle PQRS shown below if angle PSM = 70 degree then find ANgle QRM , angle PMQ and angle MRS
Answer» $\huge{\bf{\underline{\red{Solution :}}}}$ ⇒We KNOW that the diagonals of a rectangle bisect each other and both diagonal are equal in length . In ΔSPM , MS = MP $\implies \bold{ \angle PSM = \angle SPM = 70^{\circ}}$ Therefore , $\implies \bold{\angle PMS = 180^{\circ} - ( \angle PSM + \angle SPM)}$ $\implies \bold{ \angle PMS = 180^{\circ} - (70^{\circ} + 70^{\circ})}$ $\implies \bold{\angle PMS = 180^{\circ} - 140^{\circ}}$ $\implies \bold{\angle PMS = 40^{\circ}}$ Now ∠PMQ and ∠PMS from a linear pair :- $\implies \bold{\angle PMQ = 180^{\circ} - 40^{\circ}}$ $\implies \boxed{\bold{\angle PMQ = 140^{\circ}}}$ $\implies \bold{\angle PSR = \angle PSM + \angle MSR}$ $\implies \bold{90^{\circ} = 70^{\circ} + \angle MSR}$ $\implies \bold{\angle MSR = 90^{\circ} - 70^{\circ}}$ $\implies \bold{\angle MSR = 20^{\circ}}$ In ΔMSR , MS = MR $\implies \bold{\angle MSR = \angle MRS}$ $\implies \boxed{\bold{\angle MRS = 20^{\circ}}}$ Now FIND ∠QRM :- $\implies \bold{\angle QRS = \angle QRM + \angle MRS}}$ $\implies \bold{90^{\circ} = \angle QRM + \angle 20^{\circ}}$ $\implies \boxed{\bold{\angle QRM = 70^{\circ}}}$ $\rule{200}2$