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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
If $\sin \alpha=\frac{1}{\sqrt{5}}$ and $\sin \beta=\frac{3}{5}$, then $\beta-\alpha$ lies the interval
  • ✓
    $\left(0, \frac{\pi}{4}\right)$
  • B
    $\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$
  • C
    $[0, \pi]$
  • D
    $\left(\pi, \frac{5 \pi}{4}\right)$
✓

Answer

Correct option: A.
$\left(0, \frac{\pi}{4}\right)$
(A)
We have, $\sin \alpha=\frac{1}{\sqrt{5}}$
$\therefore \cos \alpha=\sqrt{1-\left(\frac{1}{\sqrt{5}}\right)^2}=\frac{2}{\sqrt{5}}$and $\sin \beta=\frac{3}{5}$
$\therefore \cos \beta=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}$
$\therefore \sin (\beta-\alpha)=\sin \beta \cos \alpha-\cos \beta \sin \alpha$
$=\frac{3}{5} \times \frac{2}{\sqrt{5}}-\frac{4}{5} \times \frac{1}{\sqrt{5}}=\frac{2}{5 \sqrt{5}}$
$=0.1789$
Now, $\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}=0.7071$
Since, $0<0.1789<0.7071$
$\therefore \sin 0<\sin (\beta-\alpha)<\sin \frac{\pi}{4}$
$\Rightarrow 0<(\beta-\alpha)<\frac{\pi}{4}$

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