Question

锐角三角形ABC中,bcosC=3,csinB=4,a=7,则sinA=?

Answer 1

由正弦定理,有：b/sinB＝c/sinC,得：bsinC＝csinB＝4,又bcosC＝3,

∴（bsinC）^2＋（bcosC）^2＝25,∴b＝5.

∴5sinC＝4,∴sinC＝4/5,

∵5cosC＝3,∴cosC＝3/5.

再由正弦定理,有：a/sinA＝b/sinB,∴asinB＝bsinA,∴7sin（180°－A－C）＝5sinA,

∴7sin（A＋C）＝5sinA,∴7sinAcosC＋7cosAsinC＝5sinA,

∴（21/5）sinA＋（28/5）cosA＝5sinA,∴28cosA＝4sinA,∴7cosA＝sinA,

∴49［1－（sinA）^2］＝（sinA）^2,∴49＝50（sinA）^2,∴sinA＝7√2/10.