A) \[{{120}^{o}}\]
B) \[{{360}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{90}^{o}}\]
Correct Answer: C
Solution :
Since, \[{{45}^{o}}\] in \[{{60}^{o}}\] Therefore, \[{{75}^{o}}\] is an isosceles \[\angle CAD={{50}^{o}}\]. Now, by the property of isosceles \[\angle BED={{120}^{o}}\]. \[\angle BCD\] Now, \[{{75}^{o}}\] (Sum of angles of\[{{105}^{o}}\]) Now, ABCD is a cyclic quadrilateral \[{{85}^{o}}\] \[{{60}^{o}}\] \[\angle BAC={{45}^{o}}\] \[\Rightarrow \]\[\angle A=\angle BAC=50{}^\circ .\]
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