Explanation:
1.[tex]\theta_{4}=90^{\circ}, \theta_{1}=\theta^{\circ}[/tex] (diagram)
2.
vector loop equation is
[tex]\overline{R_{1}}+\overline{R_{2}}+\overline{R_{3}}=\overline{R_{4}}\\[/tex]
the exponential form,
[tex]r_{1} e^{j \exp (0)}+r_{2} e^{j \theta_{2}}+r_{3} e^{j \theta_{3}}=r_{4} e^{j(90)}[/tex]
[tex]e=\cos \theta+j \sin \theta[/tex]
3.
Real part,[tex]r_{1}+\ r_{2} \cos \theta_{2}+\ r_{3} \cos \theta_{3}=0[/tex] (i)
Imaginary part,[tex]0+r_{2} \sin \theta_{2}+r_{3} \sin \theta_{3}=r_{4}[/tex] (ii)
From above equation,we get unknown variables,
Now,[tex]\theta_{2}, \theta_{3} \& r_{4}[/tex]
If [tex]\theta_{2}[/tex] is given,
From (i), [tex]\theta_{3}=\cos ^{-1}\left(\frac{-r_{1}-r_{2} \cos \theta_{2}}{r_{3}}\right)[/tex]
and from (ii), we get
[tex]x_{4}=r_{2} \sin \theta_{2}+r_{3} \sin \left[\cos ^{-1}\left(\frac{-\gamma_{1}-r_{2} \cos \theta_{2}}{r_{3}}\right)\right][/tex]
