Question

If $\sin \alpha =\frac{1}{\sqrt{5}}and\sin \beta =\frac{3}{5},\left(\alpha \text{and}\beta \text{lies in the first quadrant}\right)then\beta -\alpha$ lies in the interval

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

Answer 1

The correct option is A $[0,\frac{\pi }{4}]$

It is given that $\alpha$ and $\beta$ lies in the first quadrant.
We have $\sin \alpha =\frac{1}{\sqrt{5}}\Rightarrow \cos \alpha =\frac{2}{\sqrt{5}}$
and $\sin \beta =\frac{3}{5}\Rightarrow \cos \beta =\frac{4}{5}$
$\sin (\beta -\alpha )=\sin \beta \cos \alpha -\sin \alpha \cos \beta$
$\frac{3}{5}.\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}}.\frac{4}{5}=\frac{2}{5\sqrt{5}}=0.1789$
Now $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}=0.7071$
Since 0 < 0.1789 < 0.7071
$\therefore \sin 0$ < $\sin (\beta -\alpha )$ < $\sin \frac{\pi }{4}$ $\Rightarrow 0<(\beta -\alpha )$ < $\frac{\pi }{4}$