Q188 / 270 · NDA 2012 (Phase 2) · Mathematics

If $\sin A=\frac{2}{\sqrt{5}}$ and $\cos B=\frac{1}{\sqrt{10}}$ where $A$ and $B$ are acute angles, then what is $\mathrm{A}+\mathrm{B}$ equal to?

  • Correct answer
    A
    $135^{\circ}$
  • B
    $90^{\circ}$
  • C
    $75^{\circ}$
  • D
    $60^{\circ}$

Solution

$\sin \mathrm{A}=\frac{2}{\sqrt{5}}$ $\therefore \quad \cos A=\sqrt{1-\sin ^2 A}=\sqrt{1-\frac{4}{5}}=\frac{1}{\sqrt{5}}$ $\cos B=\frac{1}{\sqrt{10}}$ $\therefore \quad \sin \mathrm{B}=\sqrt{1-\cos ^2 \mathrm{~B}}=\frac{3}{\sqrt{10}}$ $\sin (A+B)=\sin A \times \cos B+\cos A \times \sin B$ $\begin{aligned} & =\frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{10}}+\frac{1}{\sqrt{5}} \times \frac{3}{\sqrt{10}} \\ & =\frac{2}{5 \sqrt{2}}+\frac{3}{5 \sqrt{2}}=\frac{5}{5 \sqrt{2}}=\frac{1}{\sqrt{2}}\end{aligned}$ $\begin{aligned} \sin (\mathrm{A}+\mathrm{B}) & =\sin \frac{\pi}{4} \\ \mathrm{~A}+\mathrm{B} & =45^{\circ} \text { or } 135^{\circ} .\end{aligned}$

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

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