Solution
Let the quadrilateral be
From the above
[tex]\begin{gathered} x+y+140+150=360 \\ x+y+290=360 \\ x+y=360-290 \\ x+y=70 \\ \text{Multiply through by 4} \\ 4x+4y=280\ldots\ldots\ldots\text{.}(1) \end{gathered}[/tex]Without the loss of generality, let
[tex]\begin{gathered} x\colon y=3\colon4 \\ \frac{x}{y}=\frac{3}{4} \\ \text{cross multiply} \\ 4x=3y \end{gathered}[/tex]Substitute 4x = 3y into equation (1)
[tex]\begin{gathered} 4x+4y=280 \\ 3y+4y=280 \\ 7y=280 \\ y=\frac{280}{7} \\ y=40 \end{gathered}[/tex]From
[tex]\begin{gathered} x+y=70 \\ x=70-y \\ x=70-40 \\ x=30 \end{gathered}[/tex]Therefore, the two angles are
[tex]30^{\circ},40^{\circ}[/tex]
