Quantitative Aptitude Ques 2239

Question: 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

Options:

A) $ 240{}^\circ $

B) $ 300{}^\circ $

C) $ 320{}^\circ $

D) $ 360{}^\circ $

Show Answer

Answer:

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 $

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