Let us,

(1) Analyze the girder for live load:

Considering the girder as a simply supported girder and taking moments about the left end A,

$\sum M_A=0$

$-V_B\times 60+(20\times 15)\times(15+\frac{15}{2})=0$

$V_B=112.5kN$

$\sum F_y=0(\uparrow +ve)$

$V_A-(20\times 15)+112.5=0$

$V_A=187.5kN$

Beam moment at C

$BM@C=112.5\times 30=3375kNm$

Horizontal reaction for the cable

$H_e=\frac{B,M_C}{h}=\frac{3375}{7.5}=450kN$

Let we be the uniformly distributed load transmitted of the cable,

$H_e=\frac{Wel^2}{8h}=\frac{W_e\times 60^2}{8\times 7.5}=450$

$\therefore W_e=7.5kN/m$

(2) Analysis for the dead load

Free B.M, for the girder at the centre due to dead load

$M'_e=\frac{Wl^2}{8}=\frac{10\times 60^2}{8}=4500kNm$

Horizontal reaction at cable end due to dead load

$H'=\frac{M'_e}{h}=\frac{4500}{7.5}=600kN$

Let w'e=uniformly distributed load transmitted to the cable

$H'=\frac{Wl^2}{8h}=600=\frac{w'e\times 60^2}{8\times 7.5}$

$w'e=w_d=10kN/m$

Vertical reaction at each end of the cable

$V=\frac{(w_e+w'_e)l}{2}=(7.5+10)\frac{60}{2}=525kN$

Horizontal reaction at each end of the cable

$H=H_c+H'=450+600=1050kN$

Maximum tension in the cable

$T_{max}=\sqrt{V^2+H^2}$

$=\sqrt{525^2+1050^2}=1173.9357kN$

$T_{max}=1173.9357kN$