Answer
The direction of the net gravitational force on sphere B is an angle of $~~60.6^{\circ}~~$ above the positive x axis.
Work Step by Step
We can find the y-component of the gravitational force on B:
$F_y = \frac{G~m_A~m_B}{d_1^2}$
$F_y = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.00~kg)(5.00~kg)}{(0.300~m)^2}$
$F_y = 1.853\times 10^{-8}~N$
We can find the x-component of the gravitational force on B:
$F_x = \frac{G~m_C~m_B}{d_2^2}$
$F_x = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.00~kg)(5.00~kg)}{(0.400~m)^2}$
$F_x = 1.042\times 10^{-8}~N$
We can find the direction of the net gravitational force on sphere B as an angle $\theta$ above the positive x axis:
$tan~\theta = \frac{F_y}{F_x}$
$tan~\theta = \frac{1.853\times 10^{-8}~N}{1.042\times 10^{-8}~N$}$
$\theta = tan^{-1}~(\frac{1.853\times 10^{-8}~N}{1.042\times 10^{-8}~N})$
$\theta = tan^{-1}~(1.778)$
$\theta = 60.6^{\circ}$
The direction of the net gravitational force on sphere B is an angle of $~~60.6^{\circ}~~$ above the positive x axis.
