Drag and drop an answer to each box to correctly complete the proof.Given: m∥nm∥n , m∠1=50∘m∠1=50∘ , and m∠2=42∘m∠2=42∘ .Prove: m∠5=92∘It is given that m∥nm∥n , m∠1=50∘m∠1=50∘ , and m∠2=42∘m∠2=42∘ . By the , m∠3=88∘m∠3=88∘ . Because angles formed by two parallel lines and a transversal are congruent, ∠3≅∠4∠3≅∠4 . By the angle congruence theorem, m∠3=m∠4m∠3=m∠4 . Using substitution, 88∘=m∠488∘=m∠4 . Angles 4 and 5 form a linear pair, so by the , m∠4+m∠5=180∘m∠4+m∠5=180∘ . Substituting gives 88∘+m∠5=180∘88∘+m∠5=180∘ . Finally, by the , m∠5=92∘m∠5=92∘ .
Question
Answer:
It is given that m ∥ n, m∠1 = 50° , and m∠2 = 42°. By the triangle sum theorem, m∠3 = 88°. Because corresponding angles formed by two parallel lines and a transversal are congruent, ∠3 ≅ ∠4. By the angle congruence theorem, m∠3 =m∠4. Using substitution, 88°=m. Angles 4 and 5 form a linear pair, so by the linear pair postulate, m∠4 + m∠5=180°. Substituting gives 88° + m∠5=180°. Finally, by the subtraction property of equality, m∠5 = 92°.
solved
general
11 months ago
6008