In ΔXYZ, points P, Q and R are points on the sides XY, YZ and XZ respectively ∠YXZ = 50 degree, XR = PR, ZR = QR and ∠RQZ = 80 degree. What is the value of ∠PRQ?

80 degree

In \(\Delta \)XPR,

\(\angle\)PXR = \({50}^\circ\)

PR = XR

Therefore, triangle is isosceles

\(\angle\)XPR = \(\angle\)PXR = 50

\(\angle\)XPR = 180 - 50 - 50

\(\angle\)XPR = \({80}^\circ\)

Similarly,

= \(\angle\)RQZ = \(\angle\)RZQ = 80

= \(\angle\)QRZ = 180 - 80 - 80

= \(\angle\)QRZ = \({20}^\circ\)

Now,

\(\angle\)PRQ = 180 - 20 - 80 = \({80}^\circ\).

Therefore, \(\angle\)PRQ is \({80}^\circ\).