第一问——
sinB=sin(A+C)=sinAcosC+cosAsinC=√5cosC,已知cosA,可求sinA,分别将sinA和cosA带入这个等式,等式里就剩下sinC和cosC了,这样可求tanC,第一问完毕。
第二问——
前面求得的tanC应该是√5
我觉得应该这么算:
(sinC)^2=(sinC)^2/[(sinC)^2+(cosC)^2],分子分母同时除以(cosC)^2,可得:
(sinC)^2=(tanC)^2/[(tanC)^2+1],求出sinC的平方后,开平方肯定为正,所以可以求得sinC,求得了这个,剩下就好办了,就这样。
(1)
cosA=2/3 => sinA = √5/3
sinB=√5cosC
sin(A+C)=√5cosC
sinA.cosC + cosA.sinC =√5cosC
(√5/3)cosC + (2/3)sinC =√5cosC
(√5/3)+ (2/3)tanC =√5
tanC = (2/3)√5 /(2/3)
= √5
(2)
tanC =√5 =>sinC =√(5/6), cosC = 1/√6
sinB=√5cosC
=(√5) (1/√6)
=√(5/6)
a/sinA = b/sinB
√2/(√5/3) = b/√(5/6)
b=√3
SΔABC
=(1/2)absinC
=(1/2)(√2)(√3)√(5/6)
=√5/2
希望有用
我会做