在三角形ABC中，B=120度，a=3，c=5,则sinA+sinC的值为
展开全部
By cosine-rule
b^2 =a^2 +c^2 -2abcosB
=9+25 + 15
= 49
b=7
By sine-rule
a/sinA =b/sinB = c/sinC
sinA = asin120度/b = 3√3/(2b)
sinC = csin120度/b = 5√3/(2b)
sinA+sinC = 4√3/(b) = 4√3/7
b^2 =a^2 +c^2 -2abcosB
=9+25 + 15
= 49
b=7
By sine-rule
a/sinA =b/sinB = c/sinC
sinA = asin120度/b = 3√3/(2b)
sinC = csin120度/b = 5√3/(2b)
sinA+sinC = 4√3/(b) = 4√3/7
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
