Given:
[tex]\begin{gathered} m\text{ }\angle RST=(3x-2)^0 \\ m\text{ }\angle TSK=(2x+22)^0 \end{gathered}[/tex]
The angles form a linear pair.
Required:
[tex]m\angle RST\text{ = ?}[/tex]
From the linear pair theorem:
The Linear Pair Theorem states that two angles that form a linear pair are supplementary; that is, their measures add up to 180 degrees.
Applying this theorem, we have:
[tex]m\text{ }\angle RST\text{ +m }\angle TSK\text{ = 180}[/tex]
Substituting we have:
[tex]\begin{gathered} (3x-2)\text{ + (2x + 22) = }180 \\ \text{Collect like terms} \\ 3x\text{ + 2x -2 + 22 = 180} \\ 5x\text{ = 180-20} \\ 5x\text{ = 160} \\ \text{Divide both sides by }5 \\ \frac{5x}{5}=\frac{160}{5} \\ x\text{ = 32} \end{gathered}[/tex]
The required angle is:
[tex]\begin{gathered} m\angle RST\text{ = 3x -2} \\ \text{Substituting} \\ =\text{ 3(32)-2} \\ =\text{ }96-2 \\ =\text{ 94} \end{gathered}[/tex]
Name of the theorem: linear pair theorem:
