Question

在△ABC中,sinA:sinB=根号2:1,且c²=b²+根号2bc,求角B的度数

Answer 1

根据正弦定理：a/sinA =b/sinB → sinA /sinB=a/b

由题意可得：a/b=√2/1 → a=√2b

根据余弦定理：cosA=（b^2+c^2-a^2)/2bc

cosA=[b^2+(b^2+√2bc)-2b^2]/2bc =√2bc/2bc = √2/2

∠A=45°

sin45°/sinB=√2/1 ,sinB=1/2

∠B=30°