SOLUTION
Using the theorem
The sum of angles in a triangle is 180°
Then we have
[tex]P^0+Q^0+R^0=180^0[/tex]from the diagram in the questions
[tex]\begin{gathered} P=x-11 \\ Q=3x+6 \\ R=x \end{gathered}[/tex]Substituting the parameters we have
[tex]\begin{gathered} (x-11)^0+(3x+6)^0+x^0=180^0 \\ \text{ Remove the parenthesis} \\ x-11+3x+6+x=180^0 \\ \text{ Rearrange the expression} \\ x+3x+x+6-11=180^0 \end{gathered}[/tex]Simplify the equation above
[tex]\begin{gathered} 5x-5=180^0 \\ 5x=180+5 \\ 5x=185 \\ \text{Divide both sides by 5} \\ x=\frac{185}{5} \end{gathered}[/tex]Therefore
[tex]x=39^0^{}[/tex]Hence
The value of x is 39°