A) \[{{30}^{o}}\]
B) \[{{27}^{o}}\]
C) \[{{36}^{o}}\]
D) \[{{16}^{o}}\]
Correct Answer: B
Solution :
We have, ABCD, CEFG and CIHJ are all squares. So, \[\angle 1+\angle 2+x={{90}^{o}}\] ???(i) \[{{36}^{o}}+\angle 1+x={{90}^{o}}\]q ??.(ii) \[x+\angle 2+{{27}^{o}}={{90}^{o}}\] ??.(iii) Adding (ii) and (iii), we get \[{{36}^{o}}+x+{{27}^{o}}+(\angle A+\angle 2+x)={{180}^{o}}\] \[\Rightarrow \] \[{{63}^{o}}+x+{{90}^{o}}={{180}^{o}}\] (From (i)) \[\Rightarrow \] \[x={{180}^{o}}-{{153}^{o}}={{27}^{o}}\]You need to login to perform this action.
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