Given that ΔJKL is similar to ΔDEF.
To find the side lengths, we have:
[tex]\frac{DF}{JL}=\frac{DE}{JK}=\frac{EF}{KL}[/tex]Let's solve for y:
[tex]\begin{gathered} \frac{13}{y+8}=\frac{10}{8} \\ \\ \text{Cross multiply:} \\ 10(y+8)=8(13) \\ \\ 10y+80=104 \\ \\ 10y+80-80=104-80 \\ \\ 10y=24 \end{gathered}[/tex]Divide both sides by 10:
[tex]\begin{gathered} \frac{10y}{10}=\frac{24}{10} \\ \\ y=2.4 \end{gathered}[/tex]Let's solve for x:
[tex]\begin{gathered} \frac{x-7}{12}=\frac{10}{8} \\ \\ 8(x-7)=10(12) \\ \\ 8x-56=120 \\ \\ 8x-56+56=120+56 \\ \\ 8x=176 \\ \\ \frac{8x}{8}=\frac{176}{8} \\ \\ x=22 \end{gathered}[/tex]Since they are similar triangles the corresponding angles are congruent.
Thus, we have:
z = 77 degrees
To solve for t, use the triangle angle sum theorem, which states that the sum of interior angles in a triangle is 180 degrees.
Therefore, we have:
t + 77 + 43 = 180
t + 120 = 180
Subtract 120 from both sides:
t + 120 - 120 = 180 - 120
t = 60 degrees
ANSWER:
a) x = 22
b) y = 2.4
c) z = 77°
d) t = 60°