we have the following:
[tex]<\text{JKM}=<\text{MKL}[/tex]
Therefore:
[tex]\frac{\sin (2x-3)}{4x-1}=\frac{\sin (2x-3)}{35}[/tex]
solving for x:
[tex]\begin{gathered} \frac{2x-3}{4x-1}=\frac{2x-3}{35} \\ 35\cdot(2x-3)=(4x-1)\cdot(2x-3)_{} \\ 70x-105=8x^2-14x+3 \\ 8x^2-84x+108=0 \\ 4\mleft(2x-3\mright)\mleft(x-9\mright)=0 \\ x=9 \end{gathered}[/tex]
therefore,
[tex]<\text{JKM=2}\cdot9-3=18-3=15[/tex]
