A) \[{{32}^{{}^\circ }}\]
B) \[{{37}^{{}^\circ }}\]
C) \[{{34}^{{}^\circ }}\]
D) \[{{35}^{{}^\circ }}\]
Correct Answer: B
Solution :
(b): \[\angle BCD+\angle BCE={{180}^{{}^\circ }}\Rightarrow \angle BCD+{{75}^{{}^\circ }}={{180}^{{}^\circ }}\] \[\Rightarrow \] \[\angle BCD=({{180}^{{}^\circ }}-{{75}^{{}^\circ }})={{105}^{{}^\circ }}\] In, \[\Delta BCD\] we have : \[\angle CBD+\angle BCD+\angle BDC={{180}^{{}^\circ }}\] \[\Rightarrow \]\[{{38}^{{}^\circ }}+{{105}^{{}^\circ }}+\angle BDC={{180}^{{}^\circ }}\] \[\therefore \]\[\angle BDC={{37}^{{}^\circ }}\] Since \[AB\parallel DC\], we have, \[\angle ABD=\angle BDC={{37}^{{}^\circ }}\] \[\therefore \]\[\angle ABD={{37}^{{}^\circ }}\]You need to login to perform this action.
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