Given:
[tex]\begin{gathered} \Delta ABC \\ m\angle ABC=60^{\circ},m\angle BAC=(5x-40)^{\circ},m\angle ACB=(3x)^{\circ} \end{gathered}[/tex]
The sum of all angles of the triangle is 180 degrees.
[tex]\begin{gathered} m\angle ABC+m\angle BAC+m\angle ACB=180^{\circ} \\ 60^{\circ}+(5x-40)^{\circ}+(3x)^{\circ}=180^{\circ} \\ 8x+20=180 \\ 8x=160 \\ x=20 \end{gathered}[/tex]
It gives,
[tex]\begin{gathered} m\angle BAC=(5x-40)^{\circ}=(5\cdot20-40)^{\circ}=60^{\circ} \\ m\angle ACB=(3x)^{\circ}=(3\cdot20)^{\circ}=60^{\circ} \end{gathered}[/tex]
As all the measure of the angles of a given triangle is equal that is 60 degrees. So, It is the equilateral triangle.
Answer:
[tex]\begin{gathered} m\angle BAC=60^{\circ} \\ \text{Type: an equilateral triangle} \end{gathered}[/tex]
