आकृति 6 .36 में, `(QR)/(QS)=(QT)/(PR)` तथा `angle1= angle 2` है दर्शाइए कि `Delta PQS~ Delta TQR` है

In Fig (QR)/(QS)=(QT)/(PR)= and angle1 = angle2. show that DeltaPQS ~DeltaTQR .

In Fig. 6.36, (QR)/(QS)=(QT)/(PR) and angle 1=angle 2 .Show that trianglePQS~triangle TQR

In fig., (QR)/(QS)=(QT)/(PR) and angle1=angle2 . Show that trianglePQS~triangleTQR . .

In Fig . (QR)/(QS) = (QT)/(PR)and |__ul(1)=|__ul(2) . Show that Delta PQS ~ Delta TQR

In the given figure, (QT)/(PR) = (QR)/(QS) and angle 1 = angle 2 Prove that Delta PQS ~ Delta TQR .

In the given figure, (QR)/(QS)= (QT)/(PR) and angle1 = angle2 then prove that trianglePQS ~ triangleTQR.

In the given figure, (QT)/(P)R = (QR)/(QS) and angle 1 = angle 2 . Prove that triangle PQS ~ triangle TQR .

The given figure shows PQ = PR and angle Q = angle R . Prove that Delta PQS = Delta PRT

In Delta PQR it is given that = (PQ)/(PR) = (QS)/(SR) . If angle Q = 70^(@) and angle R = 50^(@) then angle QPS = ?