Question

In the rectnagle PQRS shown below if angle PSM = 70 degree then find ANgle QRM , angle PMQ and angle MRS

Answer 1

$\huge{\bf{\underline{\red{Solution :}}}}$

⇒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 :-

$\implies \bold{\angle QRS = \angle QRM + \angle MRS}}$

$\implies \bold{90^{\circ} = \angle QRM + \angle 20^{\circ}}$

$\implies \boxed{\bold{\angle QRM = 70^{\circ}}}$

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Answer 2

$\green{ \fcolorbox{grey}{grey}{ \checkmark \: \textsf{Verified \: answer}}}$

∠MRS=20°

∠PMQ=140°

∠QRM=70°