$\displaystyle T_1 \cos(40) = T_2 \cos(50)$
$\displaystyle T_1 = \frac{T_2 \cos(50)}{\cos(40)}$
$\displaystyle T_1 \sin(40) + T_2 \sin(50) + 150 = 400$
$\displaystyle T_1 \sin(40) + T_2 \sin(50) = 250$
$\displaystyle \frac{T_2 \cos(50)\sin(40)}{\cos(40)} + T_2 \sin(50) = 250$
$\displaystyle T_2\left[\cos(50)\tan(40) + \sin(50)] = 250$
$\displaystyle T_2 = 192 \, N$
$\displaystyle T_1 = 161 \, N$
forces on the pulley in the x-direction (leftward) ...
$\displaystyle T_2 \cos(50)$
forces on the pulley in the y-direction (downward) ...
$\displaystyle T_2 \sin(50) + T_2$
Resolving forces acting on the pulley ...
$\displaystyle |F| = \sqrt{[T_2 \cos(50)]^2 + [T_2 \sin(50) + T_2]^2} = 360 \, N$
$\displaystyle \theta = \arctan\left[\frac{ T_2 \sin(50) + T_2}{T_2 \cos(50)}\right] = 70^\circ$ relative to the horizontal ... same as $\displaystyle 20^\circ$ relative to the vertical