SIN COS <-- Need Help Badly

[gone83]

Dear all,,

How to solve this.

40 = 1500Y sin X
10 = 1500Y cos x
find X (0 - 90 degree) and Y.

Thanks in advance

regards

 

[soroban]

Hello, gone83!

\(\displaystyle \begin{array}{ccc}1500y\sin x \:=\:40 & [1] \\ 1500y\cos x \:=\:10 & [2] \end{array}\)

\(\displaystyle \text{Find }x\;(0^o \,\text{-}\, 90^o)\,\text{ and }y.\)

\(\displaystyle \text{Divide [1] by [2]: }\;\frac{1500y\sin x}{1500y\cos x} \:=\:\frac{40}{10} \quad\Rightarrow\quad \tan x \:=\:4\)

\(\displaystyle \text{Therefore: }\;x \:=\:\arctan(4) \;\approx\;\boxed{76^o}\)


\(\displaystyle \text{Substitute into [1]: }\;1500y(\sin76^o) \:=\:40 \quad\Rightarrow\quad y \:=\:\frac{40}{1400\sin76^o}\)

\(\displaystyle \text{Therefore: }\;y \:=\:0.027487371 \:\approx\:\boxed{0.027}\)

 

[Deleted member 4993]

Another way to find "y"

1500[sup:3t6c6r02]2[/sup:3t6c6r02] * y[sup:3t6c6r02]2[/sup:3t6c6r02] * sin[sup:3t6c6r02]2[/sup:3t6c6r02]x + 1500[sup:3t6c6r02]2[/sup:3t6c6r02] * y[sup:3t6c6r02]2[/sup:3t6c6r02] * cos[sup:3t6c6r02]2[/sup:3t6c6r02]x = 40[sup:3t6c6r02]2[/sup:3t6c6r02] + 10[sup:3t6c6r02]2[/sup:3t6c6r02]

1500[sup:3t6c6r02]2[/sup:3t6c6r02] * y[sup:3t6c6r02]2[/sup:3t6c6r02] = 1700

y[sup:3t6c6r02]2[/sup:3t6c6r02] = 17/150[sup:3t6c6r02]2[/sup:3t6c6r02]

y = (?17)/150 (y is positive because sin(x) and cos(x) are positive)