Quantitative Aptitude Ques 1351
Question: In $ \Delta ABC, $ $ \angle A:\angle B:\angle C=2:3:4, $ a line CD drawn perpendicular to AB, then $ \angle ACD $ is
Options:
A) $ 80{}^\circ $
B) $ 20{}^\circ $
C) $ 40{}^\circ $
D) $ 60{}^\circ $
Show Answer
Answer:
Correct Answer: C
Solution:
- Let the angles be $ 2x, $ $ 3x $ and $ 4x. $ Since, the sum of interior angles of triangle is $ 180{}^\circ . $ Then, $ 2x+3x+4x=180{}^\circ $
$ \Rightarrow $ $ 9x=180{}^\circ $
$ \therefore $ $ x=20{}^\circ $
Now, $ \angle A=2x=2\times 20{}^\circ =40{}^\circ $
$ \angle B=3x=3\times 20{}^\circ =60{}^\circ $
$ \angle C=4x=4\times 20{}^\circ =80{}^\circ $
Now, $ AB\bot CD $ and AC be the transversal.
Then, $ \angle BCD=\angle ACD $ [alternate interior angles]
$ \therefore $ $ \angle ACD=40{}^\circ $
BETA
