Q175 / 270 · NDA 2012 (Phase 1) · Mathematics

Let $\sin (A+B)=1$ and $\sin (A-B)=1 / 2$, where $A, B$ $\in\left[0, \frac{\pi}{2}\right]$. What is the value A?

  • A
    $\frac{\pi}{6}$
  • Correct answer
    B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{8}$

Solution

$\begin{aligned} \sin (A+B) & =1 \\ \sin (A+B) & =\sin 90^{\circ} \\ A+B & =90^{\circ} \\ \sin (A-B) & =\frac{1}{2}=\sin 30^{\circ} \\ A-B & =30^{\circ} \\ 2 A & =120 \\ A & =60 \\ B & =30\end{aligned}$

Hence, the correct answer is option: 'B'.

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