Question

If $\sin \alpha =\frac{1}{\sqrt{5}}$, and $\sin \beta =\frac{3}{5}$, then $\beta -\alpha$ lies in the interval.

• A. $(0,\frac{\pi }{4})$
• B. $[\frac{\pi }{2},\frac{3\pi }{4}]$
• C. $(\frac{3\pi }{4},\pi )$
• D. $[\pi ,\frac{5\pi }{4}]$

Answer 1

Answer: (A), (C)

The correct options are
A $(0,\frac{\pi }{4})$
C $(\frac{3\pi }{4},\pi )$

We know from the pythagorian triplet 3,4,5 that $\sin {37}^o=\frac{3}{5}$
Also, $\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}<\frac{3}{5}$
Thus, the angle ${\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)\in (0,{37}^o)$
Hence $\beta -\alpha \in (0,\frac{\pi }{4})$ as in option A.
The other possibility of $\alpha$ is $\pi -{37}^o$ and similarly for $\beta .$
Hence, the range becomes $(\pi -\frac{\pi }{4},\pi -0)$ i.e. $(\frac{3\pi }{4},\pi )$ as in option C.