The Direction of magnetic field due to both the wires AO and OC are out of the page of  paper due to wire OC.

$B=\frac{{\mu }_{0}i}{4\pi d}[\sin {\varphi }_1+Sin{\varphi }_2]=\frac{{\mu }_{0}i}{4\pi d}[Sin(90-\theta /2)+\sin {90}^0] \because d=rSin\theta /2$

$B=\frac{{\mu }_{0}i}{4\pi rS\mathrm{in}\varphi /2}[1+\cos \varphi /2]$

Due to both the wires, the net magnetic field is $B=2\times \frac{{\mu }_{0}i}{4\pi rS\mathrm{in}\varphi /2}[1+\cos \varphi /2]$

$B=\frac{{\mu }_{0}i[1+\cos \varphi /2]}{2\pi rS\mathrm{in}\varphi /2}$