To answer this question we will use the following diagram as reference:
Notice that the angles that measures 80 degrees and a degrees are vertical angles, therefore:
[tex]a^{\circ}=80^{\circ}.[/tex]
Also, notice that the angles that measures b degrees and (6x+30) degrees are supplementary angles, therefore:
[tex]b^{\circ}+(6x+30)^{\circ}=180^{\circ}\text{.}[/tex]
Solving the above equation for b degrees we get:
[tex]\begin{gathered} b^{\circ}+(6x+30)^{\circ}-(6x+30)^{\circ}=180^{\circ}-(6x+30)^{\circ}, \\ b^{\circ}=180^{\circ}-(6x+30)^{\circ}\text{.} \end{gathered}[/tex]
Now, recall that the interior angles of a triangle add up to 180 degrees, therefore:
[tex]50^{\circ}+a^{\circ}+b^{\circ}=180^{\circ}\text{.}[/tex]
Substituting a degrees and b degrees we get:
[tex]50^{\circ}+80^{\circ}+180^{\circ}-(6x+30)^{\circ}=180^{\circ}\text{.}[/tex]
Adding like terms we get:
[tex]310^{\circ}-(6x+30)^{\circ}=180^{\circ}.[/tex]
Subtracting 180 degrees we get:
[tex]\begin{gathered} 310^{\circ}-(6x+30)^{\circ}-180^{\circ}=180^{\circ}-180^{\circ}, \\ 130^{\circ}-(6x+30)^{\circ}=0^{\circ}. \end{gathered}[/tex]
Therefore:
[tex]130-(6x+30)=0.[/tex]
Applying the distributive property we get:
[tex]130-6x-30=0.[/tex]
Adding 6x to the above equation we get:
[tex]\begin{gathered} 130-6x-30+6x=0+6x, \\ 100=6x\text{.} \end{gathered}[/tex]
Dividing the above equation by 6 we get:
[tex]\begin{gathered} \frac{100}{6}=\frac{6x}{6}, \\ x=\frac{50}{3}\text{.} \end{gathered}[/tex]
Answer:
(a)
[tex]x=\frac{50}{3}\text{.}[/tex]
(b)
1) Vertical angles.
2) Supplementary angles.
3) The interior angles of a triangle add up to 180 degrees.
