#### Answer

$(2x+7)^o=63^o
,\\
(x+61)^o=89^o
,\\
\left( x \right)^o=28^o$

#### 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}
(2x+7)^o
,\\
(x+61)^o
,\\
\left( x \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}
(2x+7)+(x+61)+\left( x \right)=180
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
2x+x+x=180-7-61
\\\\
4x=112
\\\\
x=\dfrac{112}{4}
\\\\
x=28
.\end{array}
Substituting $x=
28
$ in the angle $
(2x+7)^o
$ results to
\begin{array}{l}\require{cancel}
(2\cdot28+7)^o
\\\\=
(56+7)^o
\\\\=
63^o
.\end{array}
Substituting $x=
28
$ in the angle $
(x+61)^o
$ results to
\begin{array}{l}\require{cancel}
(28+61)^o
\\\\=
89^o
.\end{array}
Substituting $x=
28
$ in the angle $
(x)^o
$ results to
\begin{array}{l}\require{cancel}
28^o
.\end{array}
Hence, the measures of the angles of the triangle are
\begin{array}{l}\require{cancel}
(2x+7)^o=63^o
,\\
(x+61)^o=89^o
,\\
\left( x \right)^o=28^o
.\end{array}