第2个回答 2022-04-20
sin²A-cos²B + √2sinAsinB=-cos²C
sin²A-(1-sin²B) + √2sinAsinB=-(1-sin²C)
sin²A-1 + sin²B + √2sinAsinB=-1 + sin²C
sin²A + sin²B - sin²C=-√2sinAsinB
由正弦定理:(a/2R)² + (b/2R)² - (c/2R)²=-√2·(a/2R)·(b/2R)
a²/4R² + b²/4R² - c²/4R²=-√2ab/4R²
即:a² + b² - c²=-√2ab
∵a和b都不为零
∴两边同除以2ab:(a²+b²-c²)/2ab=-√2ab/2ab
cosC=-√2/2
∴C=3π/4