#### Answer

$(3x+7)^o=40^o
,\\
(9x-4)^o=95^o
,\\
\left( 4x+1 \right)^o=45^o$

#### 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}
(3x+7)^o
,\\
(9x-4)^o
,\\
\left( 4x+1 \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}
(3x+7)+(9x-4)+\left( 4x+1 \right)=180
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
3x+9x+4x=180-7+4-1
\\\\
16x=176
\\\\
x=\dfrac{176}{16}
\\\\
x=11
.\end{array}
Substituting $x=
11
$ in the angle $
(3x+7)^o
$ results to
\begin{array}{l}\require{cancel}
(3\cdot11+7)^o
\\\\=
(33+7)^o
\\\\=
40^o
.\end{array}
Substituting $x=
11
$ in the angle $
(9x-4)^o
$ results to
\begin{array}{l}\require{cancel}
(9\cdot11-4)^o
\\\\=
(99-4)^o
\\\\=
95^o
.\end{array}
Substituting $x=
11
$ in the angle $
(4x+1)^o
$ results to
\begin{array}{l}\require{cancel}
(4\cdot11+1)^o
\\\\=
(44+1)^o
\\\\=
45^o
.\end{array}
Hence, the measures of the angles of the triangle are
\begin{array}{l}\require{cancel}
(3x+7)^o=40^o
,\\
(9x-4)^o=95^o
,\\
\left( 4x+1 \right)^o=45^o
.\end{array}