The ICR can be located as follow:
i) draw a perpendicular to the line of action of P
ii) extend OQ to meet the perpendicular.
iii) The intersection gives ICR
In $∆OPQ,$ by sine rule
$50 / \sin 60 = 10 / \sinθ \\ Θ = 10$
From the figure,
$\ltPIQ = 30 ; \ltIPQ= 80 ; \ltPQI = 70 $
From $∆IPQ,$ by sine rule
$IP / \sin70 = IQ/ \sin80 = 50/\sin30 \\ IP = 94cm \\ IQ = 98.5cm \\ W1= 2πN/60 \\ = 2π \times 310/60 \\ = 32.5 rad/s \\ Vq = OQ \times W1 = 10 \times 32.5 = 325 cm/s \\ Also,\space Vq = IQ \times W2 \\ 325 = 98.5 \times w2 \\ W2 = 3.3 rad/s$ (anticlockwise)
(angular velocity of the connecting rod)
$Vp = IP \times w2 \\ = 94 \times 3.3 = 310.2 cm/s →$