Question

求助一道高中三角函数的题目

三角形ABC，已知a=√3，c=（√6+√2）/2，B=45度，求角A

Answer 1

用余弦定理求出b，再用正弦定理求出sinA，即可

Answer 2

by sine-rule
a/sinA =c/sinC
a/碧兆顷sinA = c/sin(π-A-B)
=c/sin(3π/4-A)
√3/sinA = (√6+√猜岩2)/[2sin(3π/4-A)]
2√3sin(3π/4-A) =(√6+√2)sinA
2√3[ (√2/2)cosA+ (√2/悔陆2)sinA) = (√6+√2)sinA
√6cosA = √2sinA
tanA = √3
A=π/3