A charge of +5.0 nC is at the corner A of an equilateral triangle ABP of side 0.20 m…

Step 1. The size of each field. $E_0 = \dfrac{kq}{r^2} = \dfrac{9.0 \times 10^{9} \times 5.0 \times 10^{-9}}{(0.20)^2}\ \text{N/C} = 1125\ \text{N/C}$.

Step 2. Opposite charges. The field of $+5.0$ nC at P points away from A. The field of $-5.0$ nC points towards B. The two arrows lie on opposite sides of AB, and each makes $60^\circ$ with it. The parts across AB cancel, and the parts along AB add: $E = 2E_0\cos 60^\circ = E_0 = 1125\ \text{N/C}$. It is parallel to AB, from A towards B.

Step 3. Equal charges. Both fields point away from their charges. The parts along AB cancel, and the parts across it add: $E = 2E_0\cos 30^\circ = 2 \times 1125 \times 0.866\ \text{N/C} = 1949\ \text{N/C}$. It points directly away from the middle of AB.