Answer
$(10x-20)^o=130^o
,\\
(x+15)^o=30^o
,\\
\left( x+5 \right)^o=20^o$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Equate the sum of the angles to $180$ and solve for $x.$ Then substitute the value of $x$ in the following given angles:
\begin{array}{l}\require{cancel}
(10x-20)^o
,\\
(x+15)^o
,\\
\left( x+5 \right)^o
.\end{array}
$\bf{\text{Solution Details:}}$
Since the sum of the angles of a triangle is always $180^o,$ then equate the sum of the given angles to $180.$ That is,
\begin{array}{l}\require{cancel}
(10x-20)+(x+15)+\left( x+5 \right)^o
=180
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
10x+x+x=180+20-15-5
\\\\
12x=180
\\\\
x=\dfrac{180}{12}
\\\\
x=15
.\end{array}
Substituting $x=
15
$ in the angle $
(10x-20)^o
$ results to
\begin{array}{l}\require{cancel}
(10\cdot15-20)^o
\\\\=
(150-20)^o
\\\\=
130^o
.\end{array}
Substituting $x=
15
$ in the angle $
(x+15)^o
$ results to
\begin{array}{l}\require{cancel}
(15+15)^o
\\\\=
30^o
.\end{array}
Substituting $x=
15
$ in the angle $
(x+5)^o
$ results to
\begin{array}{l}\require{cancel}
(15+5)^o
\\\\=
20^o
.\end{array}
Hence, the measures of the angles of the triangle are
\begin{array}{l}\require{cancel}
(10x-20)^o=130^o
,\\
(x+15)^o=30^o
,\\
\left( x+5 \right)^o=20^o
.\end{array}
