Let us solve for the third angle which is measure of angle C
[tex]\angle A+\angle B+\angle C=180^0\text{ \lparen Sum of angles in a triangle is 180}^0)[/tex]
Given:
[tex]\begin{gathered} \angle A=32^0 \\ \angle B=58^0 \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} 32^0+58^0+\angle C=180^0 \\ 90^0+\angle C=180^0 \\ \angle C=180^0-90^0=90^0 \\ \therefore\angle C=90^0 \end{gathered}[/tex]
Since one of the angles is 90⁰, therefore the triangle is a Right triangle.
Hence, the correct option is Option 3.
