A) \[{{18}^{o}},{{70}^{o}}\] and \[{{92}^{o}}\]
B) \[{{92}^{o}},\,{{70}^{o}}\] and \[{{18}^{o}}\]
C) \[{{70}^{o}},{{92}^{o}}\] and \[{{18}^{o}}\]
D) \[{{70}^{o}},{{18}^{o}}\] and \[{{92}^{o}}\]
Correct Answer: C
Solution :
Given \[AB=BD\] \[\Rightarrow \] \[\angle BAD=\angle BAD={{35}^{o}}\] \[\angle b=\angle BDA+\angle BAD\] \[\Rightarrow \] \[\angle b={{35}^{o}}+{{35}^{o}}={{70}^{o}}\] Also given \[AC=CE\] \[\Rightarrow \] \[\angle CAE=\angle CEA={{46}^{o}}\] Using exterior angle property, \[\Rightarrow \] \[\angle c=\angle CAE+\angle CEA\] \[={{46}^{o}}+{{46}^{o}}={{92}^{o}}\] In \[\Delta ABC,\] \[\angle a+\angle b+\angle c={{180}^{o}}\] \[\Rightarrow \] \[\angle a={{180}^{o}}-\angle b-\angle c\] \[\Rightarrow \] \[\angle a={{180}^{o}}-{{70}^{o}}-{{92}^{o}}={{18}^{o}}\] \[\therefore \] \[\angle a={{18}^{o}},\angle b={{70}^{o}}\] and \[\angle c={{92}^{o}}\]
You need to login to perform this action.
You will be redirected in
3 sec
