If tanX=3/5 and 0<90,then find the values of
3sinX+2cosX.
pls.
tan??
2007-05-31 9:55 pm
回答 (2)
2007-05-31 10:07 pm
✔ 最佳答案
Since tan X= 3/5Then we can draw a right angle triangle with opp. side =3 and adj.side=5.
Use Pyth.theorm,
hy. side = sq.root(3^2+5^2)=sq.root(34)
so
sinX=3/sq.root(34) and
cosX=5/sq.root(34)
then
3sinX+2cosX
=3*3/sq.root(34)+2*5/sq.root(34)
=19/sq.root(34)
=19*sq.root(34)/34
2007-05-31 10:08 pm
您可畫一個直角三角形，opposite side 是 3，adjacent side 是 5, 用 Pythagoras' Theorem, 找出 hypotenuse 是 √( 3² + 5² ) = √34
那問題就簡單了，sin X = 3 / √34, cos X = 5 / √34, 因為 0 < X < 90, 按照 CAST，所有sine, cosine values 都是正數
3 sin X + 2 cos X = (3 x 3 + 2 x 5) / √34 = 19 / √34
那問題就簡單了，sin X = 3 / √34, cos X = 5 / √34, 因為 0 < X < 90, 按照 CAST，所有sine, cosine values 都是正數
3 sin X + 2 cos X = (3 x 3 + 2 x 5) / √34 = 19 / √34
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