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If $\sin \mathrm{A}=\frac{2}{\sqrt{5}}$ and $\cos \mathrm{B}=\frac{1}{\sqrt{10}}$ where $\mathrm{A}$ and $\mathrm{B}$ are acute
$\begin{array}{ll}\text { angles, then what is } \mathrm{A}+\mathrm{B} \text { equal to } ? & {[2012-\mathrm{II}]}\end{array}$
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$\begin{array}{ll}\text { angles, then what is } \mathrm{A}+\mathrm{B} \text { equal to } ? & {[2012-\mathrm{II}]}\end{array}$
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Verified Answer
The correct answer is:
$135^{\circ}$
Given, $\sin \mathrm{A}=\frac{2}{\sqrt{5}}$ and $\cos \mathrm{B}=\frac{2}{\sqrt{10}}$
$\therefore \quad \cos \mathrm{A}=\frac{1}{\sqrt{5}} \quad \sin \mathrm{B}=\frac{3}{\sqrt{10}}$
$\therefore \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$
$=\left(\frac{2}{\sqrt{15}}\right)\left(\frac{1}{\sqrt{10}}\right)+\left(\frac{1}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right)$
$=\frac{2}{\sqrt{50}}+\frac{3}{\sqrt{50}}=\frac{5}{\sqrt{50}}=\frac{5}{\sqrt{25 \times 2}}=\frac{5}{5 \sqrt{2}}=\frac{1}{\sqrt{2}}$
$\sin (\mathrm{A}+\mathrm{B})=\frac{1}{\sqrt{2}} \Rightarrow \sin (90+45)=\frac{1}{\sqrt{2}}$
$\left(\because \sin (90+45)=\cos 45^{\circ}\right)$
$\therefore \quad A=90^{\circ}$ and $B=45^{\circ}$
$\therefore \quad A+B=90^{\circ}+45^{\circ}=135^{\circ}$
$\therefore \quad \cos \mathrm{A}=\frac{1}{\sqrt{5}} \quad \sin \mathrm{B}=\frac{3}{\sqrt{10}}$
$\therefore \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$
$=\left(\frac{2}{\sqrt{15}}\right)\left(\frac{1}{\sqrt{10}}\right)+\left(\frac{1}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right)$
$=\frac{2}{\sqrt{50}}+\frac{3}{\sqrt{50}}=\frac{5}{\sqrt{50}}=\frac{5}{\sqrt{25 \times 2}}=\frac{5}{5 \sqrt{2}}=\frac{1}{\sqrt{2}}$
$\sin (\mathrm{A}+\mathrm{B})=\frac{1}{\sqrt{2}} \Rightarrow \sin (90+45)=\frac{1}{\sqrt{2}}$
$\left(\because \sin (90+45)=\cos 45^{\circ}\right)$
$\therefore \quad A=90^{\circ}$ and $B=45^{\circ}$
$\therefore \quad A+B=90^{\circ}+45^{\circ}=135^{\circ}$
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