do me a favorBC中,角A,B,C所对的边分别为a,b,c,若sinA^2+sinC^2-sinB^2=根号3sinAsinC,则角B为
问题描述：
do me a favor
BC中,角A,B,C所对的边分别为a,b,c,若sinA^2+sinC^2-sinB^2=根号3sinAsinC,则角B为
答
由正弦定理:
a/sinA = b/sinB = c/sinC = 2R
可知
sinA = a/2R,sinB = b/2R,sinC = c/2R 代入原等式
(a/2R)² + (c/2R)² - (b/2R)² = √3 (a/2R)(c/2R) 化简得到
a² + c² - b² = √3ac -①
余弦定理:cosB = (a² + c ² -b²) / (2ac) ②
将①代入② 得到
cosB = √3ac/(2ac) = √3/2
所以∠B = 30°