If sinA = 1√10 and sinB = 1√5, where A and B are

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If sinA = $\frac{1}{\sqrt{10}}$ and sinB = $\frac{1}{\sqrt{5}}$ , where A and B are positive acute angles, then A+B=

Options:

A .

π

B .

\frac{π}{2}

C .

\frac{π}{3}

D .

\frac{π}{4}

Answer: Option D
:
D

We know that sin(A + B) = sinA cosB + cosA sinB

= $\frac{1}{\sqrt{10}}$ $\sqrt{1 - \frac{1}{5}}$ + $\frac{1}{\sqrt{5}}$ $\sqrt{1 - \frac{1}{10}}$

= $\frac{1}{\sqrt{10}}$ $\sqrt{\frac{4}{5}}$ + $\frac{1}{\sqrt{5}}$ $\sqrt{\frac{9}{10}}$ = $\frac{1}{\sqrt{50}}$ (2 + 3) = $\frac{5}{\sqrt{50}}$ = $\frac{1}{\sqrt{2}}$

$\Rightarrow$ sin(A + B) = sin $\frac{π}{4}$

Hence, A + B = $\frac{π}{4}$ .

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