Question

câu b thôi ạ

Answer 1

b: \(sinC=\dfrac{sinA+sinB}{cosA+cosB}=\dfrac{2\cdot sin\left(\dfrac{A+B}{2}\right)\cdot cos\left(\dfrac{A-B}{2}\right)}{2\cdot cos\left(\dfrac{A+B}{2}\right)\cdot cos\left(\dfrac{A-B}{2}\right)}\)

=>\(sinC=\dfrac{sin\left(\dfrac{A+B}{2}\right)}{cos\left(\dfrac{A+B}{2}\right)}=\dfrac{sin\left(\dfrac{C}{2}\right)}{cos\left(\dfrac{C}{2}\right)}\)

=>\(2\cdot sin\left(\dfrac{C}{2}\right)\cdot cos\left(\dfrac{C}{2}\right)=sin\left(\dfrac{C}{2}\right)\cdot\dfrac{1}{cos\left(\dfrac{C}{2}\right)}\)

=>\(2\cdot cos^2\left(\dfrac{C}{2}\right)=1\)

=>cos(C)=0

=>góc C=90 độ

=>ΔABC vuông tại C