Answer
a. $12$; b. $50$; c. $52$; d. $75$; e. $20$; f. $40$
Work Step by Step
$\textbf{a.}$ The number of sides of a triangle is $n=3$. The sum of the angles of a triangle is $$S=180(3-2)=180.$$ We determine $x$: $$\begin{align*}
30+x+9x+30&=180\\
10x+60&=180\\
10x&=180-60\\
10x&=120\\
x&=12.
\end{align*}$$ We determine the triangle's angles: $$\begin{align*}
30^\circ\\
(30+12)^\circ=42^\circ\\
9(12^\circ)=108^\circ
\end{align*}$$
$\textbf{b.}$ The number of sides of a triangle is $n=3$. The sum of the angles of a triangle is $$S=180(3-2)=180.$$ We determine $x$: $$\begin{align*}
x+10+x+20+50&=180\\
2x+80&=180\\
2x&=180-80\\
2x&=100\\
x&=50.
\end{align*}$$ We determine the triangle's angles: $$\begin{align*}
50^\circ\\
(50+10)^\circ=60^\circ\\
(50+20)^\circ=70^\circ
\end{align*}$$
$\textbf{c.}$ The number of sides of a quadrilateral is $n=4$. The sum of the angles of a quadrilateral is $$S=180(4-2)=360.$$ We determine $x$: $$\begin{align*}
50+x+2x+20+2x+30&=360\\
5x+100&=360\\
5x&=360-100\\
5x&=260\\
x&=52.
\end{align*}$$ We determine the quadrilateral's angles: $$\begin{align*}
50^\circ\\
(2(52)+30)^\circ=134^\circ\\
(2(52)+20)^\circ=124^\circ\\
52^\circ.
\end{align*}$$
$\textbf{d.}$ The number of sides of a quadrilateral is $n=4$. The sum of the angles of a quadrilateral is $$S=180(4-2)=360.$$ We determine $x$: $$\begin{align*}
x+x+42+x+35+x-17&=360\\
4x+60&=360\\
4x&=360-60\\
4x&=300\\
x&=75.
\end{align*}$$ We determine the quadrilateral's angles: $$\begin{align*}
75^\circ\\
(75+42)^\circ=117^\circ\\
(75+35)^\circ=110^\circ\\
(75-17)^\circ=58^\circ.
\end{align*}$$
$\textbf{e.}$ The number of sides of a pentagon is $n=5$. The sum of the angles of a pentagon is $$S=180(5-2)=540.$$ We determine $x$: $$\begin{align*}
540&=5x+2+3x+5+8x+8\\
&\phantom{}+4x+15+5x+10\\
540&=25x+40\\
540-40&=25x\\
500&=25x\\
x&=20.
\end{align*}$$ We determine the pentagon's angles: $$\begin{align*}
(5(20)+2^\circ&=102^\circ\\
(3(20)+5)^\circ&=65^\circ\\
(8(20)+8)^\circ&=168^\circ\\
(5(20)+10)^\circ&=110^\circ\\
(4(20)+15)^\circ&=95^\circ.
\end{align*}$$
$\textbf{f.}$ The number of sides of a hexagon is $n=6$. The sum of the angles of a hexagon is $$S=180(6-2)=720.$$ We determine $x$: $$\begin{align*}
720&=2(3x+16)+2x+8\\
&\phantom{}+4x-18+3x-7\\
&\phantom{}+2x+25\\
720&=17x+40\\
720-40&=17x\\
680&=17x\\
x&=40.
\end{align*}$$ We determine the hexagon's angles: $$\begin{align*}
(3(40)+16^\circ&=136^\circ\\
(3(40)+16^\circ&=136^\circ\\
(2(40)+8)^\circ&=88^\circ\\
(4(40)-18)^\circ&=142^\circ\\
(3(40)-7)^\circ&=113^\circ\\
(2(40)+25)^\circ&=105^\circ.
\end{align*}$$
