| The sides BC, CA and AB of \[\Delta ABC\] have been produced to D, E and F, respectively as shown in the figure, forming exterior angles \[\angle ACD,\]\[\angle BAE\]and \[\angle CBF.\] Then \[\angle ACD+\angle BAE+\angle CBF\]is equal to |
A) \[240{}^\circ \]
B) \[300{}^\circ \]
C) \[320{}^\circ \]
D) \[360{}^\circ \]
Correct Answer: D
Solution :
| Clearly, \[\angle 1+\angle BAE=180{}^\circ \] |
| \[\angle 2+\angle CBF=180{}^\circ \] |
| \[\angle 3+\angle ACD=180{}^\circ \] |
| \[\therefore \]\[(\angle 1+\angle 2+\angle 3)+(\angle BAE+\angle CBF+\angle ACD)=540{}^\circ \] |
| \[\Rightarrow \]\[180{}^\circ +(\angle BAE+\angle CBF+\angle ACD)=540{}^\circ \] |
| \[\Rightarrow \] \[\angle ACD+\angle BAE+\angle CBF=360{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec
