| In the figure shown, If ABCD is a square and \[\Delta AEB\] is quadrilateral. The measure of \[\angle DEC\]is |
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A) \[15{}^\circ \]
B) \[20{}^\circ \]
C) \[30{}^\circ \]
D) \[45{}^\circ \]
Correct Answer: C
Solution :
[c] We know that, AE = AD So, \[\angle AED=\angle ADE\]  \[=\frac{180{}^\circ -(90{}^\circ +60{}^\circ )}{2}=15{}^\circ \] \[(\therefore \,\angle EAB+\angle DAB=\angle EAD)\] Now, \[\angle DEC=60{}^\circ -(15{}^\circ +15{}^\circ )\] \[=30{}^\circ \] |
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