Question

Xác định dạng của ΔABC biết

\(\sqrt{sinA}+\sqrt{sinB}=2\sqrt{cos\dfrac{C}{2}}\)

Answer 1

\(\Leftrightarrow4cos\dfrac{C}{2}=\left(\sqrt{sinA}+\sqrt{sinB}\right)^2\le2\left(sinA+sinB\right)\)

\(\Leftrightarrow2cos\dfrac{C}{2}\le2sin\dfrac{A+B}{2}cos\dfrac{A-B}{2}\)

\(\Leftrightarrow cos\dfrac{C}{2}\le cos\dfrac{C}{2}.cos\dfrac{A-B}{2}\)

\(\Leftrightarrow cos\dfrac{A-B}{2}\ge1\)

\(\Leftrightarrow cos\dfrac{A-B}{2}=1\)

\(\Leftrightarrow A=B\)

\(\Rightarrow\) Tam giác cân tại C