If sin A = frac{1}{sqrt{10}} and sin B = frac | vedclass

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(B) Given that $A$ and $B$ are positive and acute angles $(0 < A, B < 90^{\circ})$.Since $\sin A = \frac{1}{\sqrt{10}}$,then $\cos A = \sqrt{1 - \sin^

Vedclass · Accepted Answer

$\frac{\pi}{4}$

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(B) Given that $A$ and $B$ are positive and acute angles $(0 Since $\sin A = \frac{1}{\sqrt{10}}$,then $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.Since $\sin B = \frac{1}{\sqrt{5}}$,then $\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.Using the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$:$\sin(A + B) = \left(\frac{1}{\sqrt{10}}\right) \left(\frac{2}{\sqrt{5}}\right) + \left(\frac{3}{\sqrt{10}}\right) \left(\frac{1}{\sqrt{5}}\right)$$= \frac{2}{\sqrt{50}} + \frac{3}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.Since $\sin(A + B) = \frac{1}{\sqrt{2}}$ and $A, B$ are acute,$A + B = \frac{\pi}{4}$.

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

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