question_answer

In the figure (not drawn to scale), ABCD is a square, ADE is an equilateral triangle and BFE is a straight line, find y.

A)  \[{{90}^{o}}\]                    

B)  \[{{45}^{o}}\]        

C)         \[{{75}^{o}}\]                 

D)         \[{{15}^{o}}\]                    

Correct Answer: C

Solution :

In \[\Delta \,AEB,\] \[\angle A=\angle DAE+\angle BAD\] \[\Rightarrow \] \[\angle A={{60}^{o}}+{{90}^{o}}={{150}^{o}}\] And,  \[AE=AB\] \[\Rightarrow \] \[\angle ABE=\angle AEB\] [Angles opposite to equal sides are equal] Now, \[\angle A+\angle ABE+\angle AEB={{180}^{o}}\]             (Angle sum property) \[\Rightarrow \]\[2\angle AEB={{180}^{o}}-{{150}^{o}}={{30}^{o}}\Rightarrow \angle AEB={{15}^{o}}\] Now, \[\angle E={{60}^{o}}\] \[\Rightarrow \] \[\angle DEF={{60}^{o}}-{{15}^{o}}={{45}^{o}}\] \[\therefore \]  In \[\Delta \,EFD,\]             \[\angle DEF+\angle EDF+\angle EFD={{180}^{o}}\] \[\Rightarrow \] \[{{45}^{o}}+{{60}^{o}}+y={{180}^{o}}\] \[\Rightarrow \] \[y={{180}^{o}}-({{45}^{o}}+{{60}^{o}})={{75}^{o}}\]