Given that
[tex]\begin{gathered} \angle K\cong\angle D \\ \angle J\cong\angle C \end{gathered}[/tex]
By the AA-similarity postulate, those triangles are similar, therefore
[tex]\angle L\cong\angle B[/tex]
Since those angles are congruent, their measures are equal.
[tex]m\angle L=m\angle B[/tex]
Using the property that states the sum of the inside angles of a triangle is equal to 180º, since we have the measures of ∠C and ∠D, we can easily calculate m∠B.
We have the following relation
[tex]\begin{gathered} m\angle C+m\angle D+m\angle B=180^o \\ \Rightarrow35^o+90^o+m\angle B=180^o \\ \Rightarrow m\angle B=180^o-35^o-90^o=55^o \end{gathered}[/tex]
Now that we have m∠B and m∠L(in terms of s), we can rewrite our previous equation as
[tex]m\angle L=m\angle B\Rightarrow3s-20=55[/tex]
Solving it for s, we have
[tex]\begin{gathered} 3s-20=55 \\ 3s=55+20 \\ 3s=75 \\ s=\frac{75}{3} \\ s=25 \end{gathered}[/tex]
And this is our result.
s = 25.
