**Given that,**

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

**We need to calculate the angle α**

**Using cosine law**

[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]

[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]

[tex]\alpha=\cos^{-1}(0.3846)[/tex]

[tex]\alpha=67.38^{\circ}[/tex]

**We need to calculate the angle β**

**Using cosine law**

[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]

[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]

[tex]\beta=\cos^{-1}(0.923)[/tex]

[tex]\beta=22.63^{\circ}[/tex]

**We need to calculate the force on 130 cm side**

**Using formula of force**

[tex]F_{130}=ILB\sin\theta[/tex]

[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]

[tex]F_{130}=0[/tex]

**We need to calculate the force on 120 cm side**

**Using formula of force**

[tex]F_{120}=ILB\sin\beta[/tex]

[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]

[tex]F_{120}=0.1385\ N[/tex]

**The direction of force is out of page.**

**We need to calculate the force on 50 cm side**

**Using formula of force**

[tex]F_{50}=ILB\sin\alpha[/tex]

[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]

[tex]F_{50}=0.1385\ N[/tex]

**The direction of force is into page.**

**Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.**