Let RA, RB, RC, RD be the relations at

A, B, C, D respectively.

Let angle made by RD with horizontal be $\theta $

Now, in triangle OPK, $\tan 65= (0.6/y) \\ y = 0.28 m $

Also, $x = 1.2 - 0.35 - y = 0.57 m $

Hence, in triangle PQR

$\cos\theta = x/(0.95) = 0.57/0.95 = 0.6 \\ \theta = 53.130 $

Now, considering $50KN$ block

Applying Lami's Theorem of equilibrium

$$50/(\sin 180-\theta) = RC/(\sin 90+ \theta) = RD/\sin 90 $$

Calculating, we get, $RC = 37.53 KN $ & $ RD = 62.52 KN$

Now, considering $100KN$ block

Applying conditions of equilibrium,

$\sum Fx= RA \sin 50 - RD\cos 53.130 =0 \\ RA = 48.95 KN \\ \sum FY= RA\cos50 - RD\sin 53.130 + RB - 100 = 0 \\ RB = 118.51 KN $

Reactions at contact points A, B, C, D are $48.95 KN, 118.51 KN, 37.5 KN, 62.5 KN $ respectively