Quantitative Aptitude Ques 1228
Question: In $ \Delta ABC, $ $ \angle A=90{}^\circ , $ $ \angle C=55{}^\circ , $ $ AD\bot BC. $ What is the value of $ \angle BAD $ ?
Options:
A) $ 60{}^\circ $
B) $ 45{}^\circ $
C) $ 55{}^\circ $
D) $ 35{}^\circ $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \because $ $ AD\bot BC, $ $ \angle ADC=\angle ADB=90{}^\circ $
$ \therefore $ $ \angle B=180{}^\circ -(90{}^\circ -55{}^\circ )=35{}^\circ $ [by sum property of triangle] Now, in $ \Delta ADC, $ the sum of three angles of a triangle is $ 180{}^\circ . $
$ \therefore $ $ \angle ADC+\angle ACD+\angle DAC=180{}^\circ $
$ \Rightarrow $ $ \angle DAC=180{}^\circ -90{}^\circ -55{}^\circ =35{}^\circ $ $ \angle BAD=90{}^\circ -35{}^\circ =55{}^\circ $
BETA
