It is given that sinA = 3/5 and cos(A+B) = -7/25 , where A and B are acute angles. Without solving for A and B, find the value of sinB.
求救....麻煩列出steps,另外,只可用compound angle formula...又或者其他中四應該學到既野(代數等等...).....= =麻煩各位了
AMA題....求救
2008-03-06 8:17 am
回答 (2)
2008-03-06 8:43 am
✔ 最佳答案
cos(A+B) =cosAcosB - sinAsinBwe have cosA = 4/5
and cos^2 B+sin^2 B=1
(4/5)cosB - (3/5)sinB = -7/25
20cosB - 15sinB = -7
let y=sinB
20 sqrt(1-y^2) - 15y= -7
400(1-y^2) = (15y-7)^2 = 225y^2 - 210y + 49
625y^2 - 210y - 351 = 0
by quadratic formular
y = [210 + 960]/1250 or [210-960]/1250] (rejected since sinB > 0)
so we have done.
y = 117/125
2008-03-06 9:16 am
002做得好!!!就係依個quadratic formula!!!!!我同學都係咁做ga!
ncy0916唔係唔好,不過一般人會好難先諗到sinB = sin(A+B-A) 依一步既......所以就...唔好意思......我唯有俾002分啦...因為佢個解法近我既程度多d
ncy0916唔係唔好,不過一般人會好難先諗到sinB = sin(A+B-A) 依一步既......所以就...唔好意思......我唯有俾002分啦...因為佢個解法近我既程度多d
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