We will have the next diagram
Then we can sum the forces in x and sum the forces in y
Forces in x
[tex]\sum ^{}_{}F_x=-T_1\sin (\alpha)+T_2\cos (\beta)=0[/tex][tex]\sum ^{}_{}F_x=-T_1\sin (15)+T_2\cos (35)=0[/tex]
Forces in y
[tex]\sum ^{}_{}F_y=T_1\cos (\alpha)+T_2\sin (\beta)=mg[/tex][tex]\sum ^{}_{}F_y=T_1\cos (15)+T_2\sin (35)=19.5(9.8)[/tex]
We simplify the equations found and we found the next system of equation
[tex]\begin{gathered} -T_1\sin (15)+T_2\cos (35)=0 \\ T_1\cos (15)+T_2\sin (35)=191.1 \end{gathered}[/tex]
then we isolate the T2 of the first equation
[tex]T_2=\frac{T_1\sin(15)}{\cos(35)}[/tex]
We substitute the equation above in the second equation
[tex]T_1\cos (15)+(\frac{T_1\sin(15)}{\cos(35)})\sin (35)=191.1[/tex]
we simplify
[tex]T_1(\cos (15)+\frac{\sin (15)\sin (35)}{\cos (35)})=191.1[/tex][tex]T_1(1.147)=191.1[/tex]
We isolate the T1
[tex]T_1=\frac{191.1}{1.147^{}}=166.6N[/tex]
then we can substitute the value we found in T1 in the euation with T2 isolate
[tex]T_2=\frac{(166.6)_{}\sin (15)}{\cos (35)}=52.54N[/tex]
