Answer
$B_{net}=9.47\times10^{-5}T$
$\theta=82^o$
Work Step by Step
$B_T=\frac{2\times10^{-7}(28A)}{0.06m}=9.33\times10^{-5}T$
$B_B=\frac{2\times10^{-7}(28A)}{0.10m}=5.6\times10^{-5}T$
Measure of top angle: $\theta_T$
Measure of right angle: $\theta_R$
Measure of bottom angle: $\theta_B$
$(10.0cm)^2=(6.0cm)^2 + (13.0cm)^2-2(6.0cm)(13.0cm)\cos(\theta_T)$
$\theta_T=47.7^o$
$(6.0cm)^2=(10.0cm)^2 + (13.0cm)^2-2(10.0cm)(13.0cm)\cos(\theta_B)$
$\theta_B=26.3^o$
$B_x=9.33\times10^{-5}T\cos(47.7^o)-5.6\times10^{-5}T\cos(26.3^o)=1.26\times10^{-5}T$
$B_x=9.33\times10^{-5}T\sin(47.7^o)+5.6\times10^{-5}T\sin(26.3^o)=9.38\times10^{-5}T$
$B_{net}=\sqrt{B_x^2+B_y^2}=9.47\times10^{-5}T$
$\theta=\frac{B_y}{B_x}=82^o$
