The given triangle can be shown below
Since the sum of angles in a triangle is 180 degrees it follows
[tex]\angle A+\angle B+\angle C=180^{\degree}[/tex]
It follows
[tex]\begin{gathered} \angle A+43.2+90=180^{\degree} \\ \angle A=180-90-43.2 \\ \angle A=46.8^{\degree} \end{gathered}[/tex]
From the triangle above
Using trigonometrical ratios
It follows
[tex]\tan 43.2=\frac{b}{3.2}[/tex]
This gives
[tex]\begin{gathered} b=3.2\times\tan 43.2 \\ b=3.2\times0.9391 \\ b=3.0 \end{gathered}[/tex]
Hence,
[tex]b=3.0m[/tex]
Also
[tex]\cos 43.2=\frac{3.2}{c}[/tex]
It follows
[tex]\begin{gathered} c\times\cos 43.2=3.2 \\ c=\frac{3.2}{\cos 43.2} \\ c=\frac{3.2}{0.7290} \\ c=4.4 \end{gathered}[/tex]
Hence,
[tex]c=4.4m[/tex]
Therefore, the solutions are
[tex]\angle A=46.8,\bar{AC}=3.0m,\bar{AB}=4.0m[/tex]
