In a triangle, if $$ a=\sqrt{2} $$ $$ ,b=2 $$ , $$ \sin B+\cos B=\sqrt{2} $$ , prove $$ A=\dfrac{\pi }{6} $$

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In a triangle, if $$ a=\sqrt{2} $$  $$ ,b=2 $$ , $$ \sin B+\cos B=\sqrt{2}  $$ , prove $$ A=\dfrac{\pi  }{6} $$

Uniteasy Admin · Accepted Answer

$$ \because \sin B+\cos B=\sqrt{2} $$
Square both sides, we get
$$ 1+2 \sin B \cos B=2 $$
$$ 2\sin B\cos B=1 $$
According to double angle sines identify,
$$ \sin 2B =1 $$
$$ \because 0 < B< \pi $$
$$ 	herefore B=\dfrac{\pi }{4} $$
According to Law of Sines,
$$ \dfrac{\sin A}{a}= \dfrac{\sin B}{b} \ $$
$$ \sin A=a\cdotp  \dfrac{\sin B}{b}=\sqrt{2}\cdotp \dfrac{\sin \dfrac{\pi}{4} }{2} =\dfrac{1}{2} $$ 
$$ \because a < b=2 $$
$$ 	herefore A < B =\dfrac{\pi}{4} $$
$$ A=\dfrac{\pi}{6} $$

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