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Geometry Difficulty 3.4 AMC 10/12 Find the answer Canada

In the diagram, ABCDABCD is a rectangle and △ACE\triangle ACE is right-angled at CC.Figure 0If ∠BAC=42°\angle BAC=42\degree, the measure of ∠DCE\angle DCE is

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Solution

Solution 1:

In △ABC\triangle ABC, ∠ACB=180°−∠ABC−∠BAC=180°−90°−42°=48°\angle ACB=180\degree-\angle ABC-\angle BAC=180\degree-90\degree-42\degree=48\degree.

Since ∠BCD=90°\angle BCD=90\degree, then $∠ACD=∠BCD−∠ACB=90°−48°=42°\$\angle ACD=\angle BCD-\angle ACB=90\degree-48\degree=42\degree.Since. Since ∠ACE=90°\angle ACE=90\degree,then, then ∠DCE=∠ACE−∠ACD=90°−42°=48°\angle DCE=\angle ACE-\angle ACD=90\degree-42\degree=48\degree.Solution2:Since. Solution 2: Since ∠ABC=∠BCD=90°\angle ABC=\angle BCD=90\degree,then, then ABisparallelto is parallel to DC.Thus,. Thus, ∠ACD=∠\angle ACD=\angle BAC
=42°=42\degreebyaparallellinestheorem( by a parallel lines theorem (Zpattern,alternateinteriorangles).Since pattern, alternate interior angles). Since ∠ACE=90°\angle ACE=90\degree,then, then ∠DCE=∠ACE−∠ACD=90°−42°=48°\angle DCE=\angle ACE-\angle ACD=90\degree-42\degree=48\degree.Solution3:Since. Solution 3: Since ∠BAD=90°\angle BAD=90\degree,then, then ∠CAD=∠BAD−∠BAC=90°−42°=48°\angle CAD=\angle BAD-\angle BAC=90\degree-42\degree=48\degree.In. In △\triangle ADC,, ∠ACD=180°−∠ADC−∠CAD=180°−90°−48°=42°$.\angle ACD=180\degree-\angle ADC-\angle CAD=180\degree-90\degree-48\degree=42\degree\$.

Since ∠ACE=90°\angle ACE=90\degree, then $∠DCE=∠ACE−∠ACD=90°−42°=48°$.\$\angle DCE=\angle ACE-\angle ACD=90\degree-42\degree=48\degree\$.

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