Question

cho tam giác ABC có 3 góc A,B,C thỏa \(sinA+sinB=cosA+cosB\)

tính giá trị góc C

Answer 1

\(sinA+sinB=cosA+cosB\)

\(\Leftrightarrow2sin\frac{A+B}{2}.cos\frac{A-B}{2}=2cos\frac{A+B}{2}cos\frac{A-B}{2}\)

\(\Rightarrow\left[{}\begin{matrix}cos\frac{A-B}{2}=0\\sin\frac{A+B}{2}=cos\frac{A+B}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A-B=180^0\left(l\right)\\\frac{A+B}{2}=90^0-\frac{A+B}{2}\end{matrix}\right.\)

\(\Rightarrow A+B=90^0\Rightarrow C=90^0\)