Question

已知三角形ABC的外接圆半径为R,且满足2R(sin方A-sin方C）=（根号2a-b)sinB,(根号中为2，再乘a，最后减b),

Answer 1

2.
a/sinA=b/sinB=c/sinC=2R=2√
=>a=2RsinA,b=2RsinB,c=2RsinC
2√2（sin²A-sin²C）=（a-b）sinB
=>4R²（sin²A-sin²C）=2R（a-b）sinB
=>a²-c²=(a-b)b
=>(a²+b²-c²)/2ab=1/2=cosC
=>C=60°

Answer 2

a/sinA=b/sinB=c/sinC=2R=2√2
=>a=2RsinA,b=2RsinB,c=2RsinC
2√2（sin²A-sin²C）=（a-b）sinB
=>4R²（sin²A-sin²C）=2R（a-b）sinB
=>a²-c²=(a-b)b
=>(a²+b²-c²)/2ab=1/2=cosC
=>C=60°
S△ABC=absinC/2=2RsinA*2RsinB*sinC/2
=√3(2sinAsinB)=√3[cos(A-B)-cos(A+B)]
=√3[cos(A-B)+1/2]≤3√3/2