Answer:
Δ JKL is similar to Δ ABC ⇒ D
Step-by-step explanation:
Similar triangles have equal angles in measures
In ΔABC
∵ m∠A = 15°
∵ m∠B = 120
∵ The sum of the measures of the interior angles of a Δ is 180°
∴ m∠A + m∠B + m∠C = 180°
→ Substitute the measures of ∠A and ∠B
∵ 15 + 120 + m∠C = 180
→ Add the like terms in the left side
∴ 135 + m∠C = 180
→ Subtract 135 from both sides
∴ 135 - 135 + m∠C = 180 - 135
∴ m∠C = 45°
The similar Δ to ΔABC must have the same measures of angles
If triangles ABC and JKL are similar, then
m∠A must equal m∠J
m∠B must equal m∠K
m∠C must equal m∠L
∵ m∠J = 15°
∴ m∠A = m∠J
∵ m∠L = 45°
∴ m∠C = m∠L
∵ m∠J + m∠K + m∠L = 180°
→ Substitute the measures of ∠J and ∠L
∵ 15 + m∠K + 45 = 180
→ Add the like terms in the left side
∴ 60 + m∠K = 180
→ Subtract 60 from both sides
∴ 60 - 60 + m∠K = 180 - 60
∴ m∠K = 120°
∴ m∠B = m∠K
∴ Δ JKL is similar to Δ ABC
