Question

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

Collected in the board: Trigonometry

Steven Zheng posted 3 months ago


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

\therefore 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

\therefore A < B =\dfrac{\pi}{4}

A=\dfrac{\pi}{6}

Steven Zheng posted 3 months ago

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