Step 1: Draw an image to illustrate the problem
Consider the forces along the horizontal axis.
[tex]\begin{gathered} -T_1\cos \theta_1+T_2\cos \theta_2=0 \\ \text{ therefore} \\ T_2\cos 20^0=T_1\cos 20^0 \end{gathered}[/tex][tex]\text{ Dividing both sides by }\cos 20^0[/tex][tex]\begin{gathered} \frac{T_2\cos20^0}{\cos20^0}=\frac{T_1\cos 20^0}{\cos 20^0} \\ \text{thus} \\ T_2=T_1 \end{gathered}[/tex]
Consider the forces along the vertical axis.
[tex]\begin{gathered} T_1\sin 20^0+T_2\sin 20^0-16=0 \\ T_1\sin 20^0+T_1\sin 20^0-16=0\text{ (}T_1=T_2) \\ \text{ Thus} \\ 2T_1\sin 20^0=16 \\ T_1=\frac{16}{2\sin 20^0}\approx23.39\text{ pounds} \end{gathered}[/tex]
then T₁ = 23.39 pounds
Since T₁=T₂, then T₂ = 23.39 pounds
Hence, smaller value = 23.4 pounds to one decimal place and
larger value = 23.4 pounds to one decimal place
