Exercise 57 Page 520

We want to find the measure of ∠ 6, given that m∠ 1=38. Let P be the point of intersection of the diagonals.

Let's recall the Interior Angles Theorem.

Therefore, the interior angles in △ ABC, which are ∠ 2, ∠ ABC, and ∠ 4, add to 180^(∘). m∠ 2+ m∠ ABC+ m∠ 4=180^(∘) Because we are given that ABCD is a rectangle, we know that m∠ DAB=90^(∘) and m∠ ABC= 90^(∘). By the Angle Addition Postulate, we can express m ∠ DAB as the sum of m∠ 2 and m ∠ CAD. m∠ 2+m∠ CAD=m∠ DAB ⇕ m∠ 2+38^(∘)=90^(∘) ⇕ m∠ 2= 52^(∘) By using inverse operations, we obtained that m∠ 2= 52^(∘). Now we have all values we need to calculate m∠ 4.

m∠ 2+m∠ ABC+ m∠ 4=180^(∘)

SubstituteII

m∠ 2= 52^(∘), m∠ B= 90^(∘)

52^(∘)+ 90^(∘)+m∠ 4=180^(∘)

AddTerms

Add terms

142^(∘)+m∠ 4=180^(∘)

SubEqn

LHS-142^(∘)=RHS-142^(∘)

m∠ 4=38^(∘)

Note that in a rectangle the diagonals are congruent and bisect each other. Therefore BP≅CP.

Let's recall the Base Angles Theorem.

Base Angles Theorem

If two sides in a triangle are congruent, then the angles opposite them are congruent.

Therefore ∠ 6 ≅ ∠ 4, which means they have the same measure, so m ∠ 6=38^(∘).