send me the answer
 

Solution:
Let, AB = AC = 40 cm = x



Electric field at A due to charge at B   (+q),
${\overrightarrow{E}}_{AB}=\frac{Kq}{x^2} \mathrm{along} \mathrm{B} \mathrm{to} \mathrm{A}$
Electric field at A due to charge at C   (-q),
${\overrightarrow{E}}_{AC}=\frac{Kq}{x^2} \mathrm{along} \mathrm{A} \mathrm{to} \mathrm{C}$
So, net electric field at A due to both the charges,
$$\begin{gathered} \left|{\overrightarrow{E}}_R\right|=\sqrt{{\left|{\overrightarrow{E}}_{AB}\right|}^2+{\left|{\overrightarrow{E}}_{AC}\right|}^2+2\left|{\overrightarrow{E}}_{AB}\right|\left|{\overrightarrow{E}}_{AC}\right|\cos \left(180°-30°\right)} \\ \Rightarrow \left|{\overrightarrow{E}}_R\right| =\frac{Kq }{x^2}\sqrt{1+1-2\cos 30°} \\ = \frac{Kq }{x^2}\sqrt{1+1-2\cos 30°} \\ =\frac{9\times {10}^9\times 2\times {10}^{-6}}{{\left(40\times {10}^{-2}\right)}^2}\times \sqrt{1+1-2\times \frac{\sqrt{3}}{2}} \\ =5.8\times {10}^4 \mathrm{N}/\mathrm{C} \end{gathered}$$

By symmetry, the direction of the resultant field will be at an angle $75°$ to the side AC as shown in the diagram.