A) \[{{88}^{{}^\circ }}\]
B) \[{{48}^{{}^\circ }}\]
C) \[{{118}^{{}^\circ }}\]
D) \[{{108}^{{}^\circ }}\]
Correct Answer: A
Solution :
(a): From figure, \[AB\parallel DC\] and AD is transversal \[\therefore \]\[\angle BAD=\angle EDC={{48}^{{}^\circ }}\] In \[\Delta EDC\], \[\angle EDC+\angle ECD+\angle DEC={{180}^{{}^\circ }}\] \[{{48}^{{}^\circ }}+{{44}^{{}^\circ }}+x={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\text{ }x={{180}^{{}^\circ }}-\left( {{48}^{{}^\circ }}+{{44}^{{}^\circ }} \right)={{88}^{{}^\circ }}\]
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