Given that EDC is a straight line and \[\angle AED={{36}^{o}}\] find \[\angle BAD\].
A) \[{{36}^{o}}\]
B) \[{{72}^{o}}\]
C) \[{{108}^{o}}\]
D) \[{{120}^{o}}\]
Correct Answer: C
Solution :
\[BDC=\,\,AED={{36}^{o}}\] (Corresponding s, AE BD.) \[ABD=~BDC={{36}^{o}}\] (Alternate s, AB DC) \[ADB=~ABD={{36}^{o}}\] (Base angles of isosceles, since\[AB=DC\]) \[BAD={{180}^{o}}-ABD-ADB\] (Angle sum of a triangle.) \[={{180}^{o}}-{{36}^{o}}-{{36}^{o}}\] \[={{108}^{o}}\]
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