In Fig. 6.41, if AB || DE, 2 BAC = 35º and Z CDE = 53º, find ZDCE.
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∠AED=∠BAC (alternate interior angles)
⇒∠AED=35
0
⇒∠CED=35
0
InΔDEC
∠CDE+∠DCE+∠CED=180
0
(by angle sum property)
53
0
+∠DCE+35
0
=180
0
∠DCE+88
0
=180
0
∠DCE=180
0
−88
0
∠DCE=92
0
⇒∠AED=35
0
⇒∠CED=35
0
InΔDEC
∠CDE+∠DCE+∠CED=180
0
(by angle sum property)
53
0
+∠DCE+35
0
=180
0
∠DCE+88
0
=180
0
∠DCE=180
0
−88
0
∠DCE=92
0
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