#### Answer

$(x-30)^o=60^0
,\\
(2x-120)^o=60^o
,\\
\left( \dfrac{1}{2}x+15\right)^o=60^o$

#### 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}
(x-30)^o
,\\
(2x-120)^o
,\\
\left( \dfrac{1}{2}x+15\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}
(x-30)+(2x-120)+\left( \dfrac{1}{2}x+15\right)=180
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
x+2x+\dfrac{1}{2}x=180+30+120-15
\\\\
3x+\dfrac{1}{2}x=315
\\\\
2\left( 3x+\dfrac{1}{2}x\right)=2(315)
\\\\
6x+x=630
\\\\
7x=630
\\\\
x=\dfrac{630}{7}
\\\\
x=90
.\end{array}
Substituting $x=
90
$ in the angle $
(x-30)^o
$ results to
\begin{array}{l}\require{cancel}
(90-30)^o
\\\\=
60^o
.\end{array}
Substituting $x=
90
$ in the angle $
(2x-120)^o
$ results to
\begin{array}{l}\require{cancel}
(2\cdot90-120)^o
\\\\=
(180-120)^o
\\\\=
60^o
.\end{array}
Substituting $x=
90
$ in the angle $
\left(\dfrac{1}{2}x+15\right)^o
$ results to
\begin{array}{l}\require{cancel}
\left(\dfrac{1}{2}\cdot90+15\right)^o
\\\\=
\left(45+15\right)^o
\\\\=
60^o
.\end{array}
Hence, the measures of the angles of the triangle are
\begin{array}{l}\require{cancel}
(x-30)^o=60^0
,\\
(2x-120)^o=60^o
,\\
\left( \dfrac{1}{2}x+15\right)^o=60^o
.\end{array}