Question

find the measure of

Answer 1

we have the following:

$<\text{JKM}=<\text{MKL}$

Therefore:

$\frac{\sin (2x-3)}{4x-1}=\frac{\sin (2x-3)}{35}$

solving for x:

$\begin{gathered} \frac{2x-3}{4x-1}=\frac{2x-3}{35} \\ 35\cdot(2x-3)=(4x-1)\cdot(2x-3)_{} \\ 70x-105=8x^2-14x+3 \\ 8x^2-84x+108=0 \\ 4\mleft(2x-3\mright)\mleft(x-9\mright)=0 \\ x=9 \end{gathered}$

therefore,

$<\text{JKM=2}\cdot9-3=18-3=15$