Here's my working:
$$ V_P = \frac{V_L}{\sqrt{3}}$$ $$ \therefore V_P = \frac{21.651kV}{\sqrt{3}} = 12.5kV$$ $$ S = 3V_PI_P$$ $$ \therefore I_p = \frac{S}{3V_p} = \frac{2400+j1800kVa}{(3)(12500)} = 80\angle36.86^\circ$$
The solution given in the answer sheet is $$80\angle-36.86^\circ$$
Is there a mistake I made? because this messes up the answers for the rest of the tut. Thanks!