ANSWER
[tex]\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}[/tex]
EXPLANATION
First, we have to make a sketch of the direction of the moments of the forces about 12m from the left in the diagram:
The sum of upward forces must be equal to the sum of downward forces. This implies that:
[tex]\begin{gathered} T_A+T_B=20+10 \\ T_A+T_B=30N \end{gathered}[/tex]
Also, the sum of clockwise moments must be equal to the counter-clockwise moments:
[tex]\begin{gathered} (20\cdot8)+(T_B\cdot4)=(T_A\cdot12) \\ 160+4T_B=12T_A \\ \Rightarrow12T_A-4T_B=160 \end{gathered}[/tex]
From the first equation, make TA the subject of the formula:
[tex]T_A=30-T_B[/tex]
Substitute that into the second equation:
[tex]\begin{gathered} 12(30-T_B)-4T_B=160 \\ 360-12T_B-4T_B=160 \\ -16T_B=160-360=-200 \\ T_B=\frac{-200}{-16} \\ T_B=12.5N \end{gathered}[/tex]
Substitute that into the equation for TA:
[tex]\begin{gathered} T_A=30-12.5 \\ T_A=17.5N \end{gathered}[/tex]
Therefore, the reaction supports at TA and TB are:
[tex]\begin{gathered} T_A=17.5N \\ T_B=12.5N \end{gathered}[/tex]
