Re: [中學] 三角函數問題
: : 請問這題要怎麼解呢
: △ABC : △ACD : △ADE = 2 : 4 : 3
: △ACD / △ABC = AD sinβ / AB sinα = 2 / 1
: △ACD / △ADE = AC sinβ / AE sinγ = 4 / 3
: 2 * 4 / 3 = 8 / 3 = (sinβ * 4) / ((2 + 4 + 3) * sinαsinγ)
: => sinβ / (sinαsinγ) = 6
原式
=cos(gamma+alpha)/[sin(gamma)*sin(alpha)]
=cot(gamma)cot(alpha)-1
3/6=EAsin(gamma)/ABsin(alpha+beta)...分定
1/2=EAsin(gamma)/ABcos(gamma)...正餘弦互換
cot(gamma)=2EA/AB,同理,cot(alpha)=7AB/2EA.
=7-1=6...ans
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