Question

31. (3) Given.
( therefore quad P R=R Q, ) then ( , angle R P Q=angle R Q P=x )
( therefore angle P R Q+angle R P Q+angle R Q P=180^{circ} )
( angle P R Q+x+x=180^{circ} )
( angle P R Q=180^{circ}-2 x )
( frac{angle P R Q}{2}=frac{2left(90^{circ}-xright)}{2} )
( angle P R S=90^{circ}-x )
In ( Delta P R S )
( angle P R S+angle P S R+angle R P S=180^{circ} )
( therefore )
( 90^{circ}-x+angle P S R+x=180^{circ} )
( angle P S R=180^{circ}-90^{circ}=90^{circ} )

# 31. If S is the midpoint of a straight line PQ and R is a point different from S, such that PR = RQ then RRB NTPC (Phase I) 2016 (1) ZPRS = 90° (2) ZQRS = 90° (3) ZPSR = 90° (4) ZQSR = 90°

Solution