Answer:
388.56km[E13degS]
Step-by-step explanation:
Δdx = Δd1x + Δd2x
Δdx = 250km [E]+ 180sin(60) [E]
Δdx = 405.88 km [E]
Δdy = Δd1y + Δd2y
Δdy = 0 + 160cos(60) [S]
Δdy = 90km [S]
Δdt = √(Δdx)^2+(Δdy)^2
Δdt = √164738.57+8100
Δdt = 398.85km
tanθ = Δdy/Δdx
tanθ = 90/388.56
θ = tan-1(90/388.56)
θ = 13.04 deg