A) \[{{140}^{{}^\circ }}\]
B) \[{{45}^{{}^\circ }}\]
C) \[{{120}^{{}^\circ }}\]
D) \[{{75}^{{}^\circ }}\]
Correct Answer: A
Solution :
(a): Draw EF parallel to both. AB and CD. Now, AB I; EF and transversal AR cuts them w k and E respectively. \[\angle BAE+\angle FEA={{180}^{{}^\circ }}\] \[\Rightarrow \]\[{{108}^{{}^\circ }}\angle 1={{180}^{{}^\circ }}\] \[\Rightarrow \] \[\angle 1={{180}^{{}^\circ }}-{{108}^{{}^\circ }}={{72}^{{}^\circ }}\] Again, \[EF\parallel CD\] and transversal CE cuts them at E and F respectively. \[\angle FEC+\angle ECD={{180}^{{}^\circ }}\] \[\Rightarrow \] \[\angle 2+{{112}^{{}^\circ }}={{180}^{{}^\circ }}\] \[\Rightarrow \] \[\angle 2={{180}^{{}^\circ }}-{{112}^{{}^\circ }}\] \[\Rightarrow \] \[\angle 2={{68}^{{}^\circ }}\] Now, \[x=\angle 1+\angle 2\] \[\Rightarrow \] \[x={{72}^{{}^\circ }}+{{68}^{{}^\circ }}={{140}^{{}^\circ }}\]You need to login to perform this action.
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