A) \[{{2}^{o}}\]
B) \[{{8}^{o}}\]
C) \[{{17}^{o}}\]
D) \[{{20}^{o}}\]
Correct Answer: A
Solution :
In \[\Delta FGC,\] \[\angle CBF={{60}^{o}}\] (Angle of equilateral triangle) \[\therefore \] \[x+{{60}^{o}}+{{92}^{o}}={{180}^{o}}\] \[\Rightarrow \] \[x={{180}^{o}}-{{152}^{o}}={{28}^{o}}\] Now, In \[\Delta \,BCF,\] \[\angle CBF={{60}^{o}}\] \[\angle FCB={{180}^{o}}-{{92}^{o}}\] (Linear pair) \[\Rightarrow \] \[\angle FCB={{88}^{o}}\] \[\therefore \] \[\angle BFC+{{88}^{o}}+{{60}^{o}}={{180}^{o}}\] (Angle sum property) \[\Rightarrow \] \[\angle BFC={{180}^{o}}-{{148}^{o}}={{32}^{o}}\] And \[\angle AFE={{90}^{o}}\] \[\Rightarrow \] \[y+{{32}^{o}}={{90}^{o}}\] \[\Rightarrow \] \[y={{90}^{o}}-{{32}^{o}}={{58}^{o}}\] \[\therefore \] \[y-2x={{58}^{o}}-2\times {{28}^{o}}={{58}^{o}}-{{56}^{o}}={{2}^{o}}\]
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