Trigonometric Identities Question 248
Question: In a triangle ABC, $ \sin A-\cos B=\cos C, $ then what is B equal to?
Options:
A) $ \pi $
B) $ \pi /3 $
C) $ \pi /2 $
D) $ \pi /4 $
Show Answer
Answer:
Correct Answer: C
Solution:
In a $ \Delta ABC, $ we have $ \sin A-\cos B=\cos C\Rightarrow \sin A=\cos B+\cos C $
$ \Rightarrow ,2\sin \frac{A}{2}.\cos \frac{A}{2} $ $ =2\cos ( \frac{B+C}{2} ).\cos ( \frac{B-C}{2} ) $ $ [\because \sin 2A=2\sin A.\cos A] $ and $ \cos B+\cos C=2\cos ( \frac{B+C}{2} ).\cos ( \frac{B-C}{2} ) $
$ \Rightarrow 2\sin \frac{A}{2}.\cos \frac{A}{2}=2\cos ( 90{}^\circ -\frac{A}{2} ).\cos ( \frac{B-C}{2} ) $ $ [ \because A+B+C=180{}^\circ \Rightarrow ( \frac{B+C}{2} )=90{}^\circ -\frac{A}{2} ] $
$ \Rightarrow 2\sin \frac{A}{2}.\cos \frac{A}{2}=2\sin \frac{A}{2}.\cos ( \frac{B-C}{2} ) $ $ [\because ,\cos (90{}^\circ -\theta )=\sin \theta ] $
$ \Rightarrow \cos \frac{A}{2}=\cos ( \frac{B-C}{2} ) $
$ \Rightarrow \frac{A}{2}=\frac{B-C}{2} $
$ \Rightarrow ,A+C=B $ ?.(i) Also, $ A+C=180{}^\circ -B $ ??..(ii) So, $ 180{}^\circ -B=B $
$ \Rightarrow 2B=180{}^\circ $
$ \therefore B=90{}^\circ $
BETA
