Trigonometric Identities Question 136
Question: If $ \sin (\alpha -\beta )=\frac{1}{2} $ and $ \cos (\alpha +\beta )=\frac{1}{2}, $ where $ \alpha $ and $ \beta $ are positive acute angles, then
Options:
A) $ \alpha =45{}^\circ ,\beta =15{}^\circ $
B) $ \alpha =15{}^\circ ,\beta =45{}^\circ $
C) $ \alpha =60{}^\circ ,\beta =15{}^\circ $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \sin (\alpha -\beta )=\frac{1}{2}=\sin 30{}^\circ \Rightarrow \alpha -\beta =30{}^\circ $ ?..(i) and $ \cos (\alpha +\beta )=\frac{1}{2}\Rightarrow \alpha +\beta =60{}^\circ $ ?..(ii) Solving (i) and (ii), we get $ \alpha =45{}^\circ $ and $ \beta =15{}^\circ $ . Trick: In such type of problems, students should satisfy the given conditions with the values given in the options. Here $ \alpha =45{}^\circ $ and $ \beta =15{}^\circ $ satisfy both the conditions.
BETA
