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Question: If \(\sin A = \frac{1}{\sqrt{10}}\)and \(\sin B = \frac{1}{\sqrt{5}},\) where A and B are positive a...

If sin⁡A=110\sin A = \frac{1}{\sqrt{10}}and sin⁡B=15,\sin B = \frac{1}{\sqrt{5}}, where A and B are positive acute angles, then A+B=A + B =

A

π\pi

B

π/2\pi/2

C

π/3\pi/3

D

π/4\pi/4

Answer

π/4\pi/4

Explanation

Solution

We know that sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A\cos B + \cos A\sin B

=1101−15+151−110= \frac{1}{\sqrt{10}}\sqrt{1 - \frac{1}{5}} + \frac{1}{\sqrt{5}}\sqrt{1 - \frac{1}{10}}

=11045+15910=150(2+3)=550=12= \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}}

⇒sin⁡(A+B)=sin⁡π4\Rightarrow \sin(A + B) = \sin\frac{\pi}{4} Hence, A+B=π4A + B = \frac{\pi}{4}