G gone83 New member Joined Jun 23, 2009 Messages 1 Jun 23, 2009 #1 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
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
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jun 23, 2009 #2 Hello, gone83! \(\displaystyle \begin{array}{ccc}1500y\sin x \:=\:40 & [1] \\ 1500y\cos x \:=\:10 & [2] \end{array}\) \(\displaystyle \text{Find }x\;(0^o \,\text{-}\, 90^o)\,\text{ and }y.\) Click to expand... \(\displaystyle \text{Divide [1] by [2]: }\;\frac{1500y\sin x}{1500y\cos x} \:=\:\frac{40}{10} \quad\Rightarrow\quad \tan x \:=\:4\) \(\displaystyle \text{Therefore: }\;x \:=\:\arctan(4) \;\approx\;\boxed{76^o}\) \(\displaystyle \text{Substitute into [1]: }\;1500y(\sin76^o) \:=\:40 \quad\Rightarrow\quad y \:=\:\frac{40}{1400\sin76^o}\) \(\displaystyle \text{Therefore: }\;y \:=\:0.027487371 \:\approx\:\boxed{0.027}\)
Hello, gone83! \(\displaystyle \begin{array}{ccc}1500y\sin x \:=\:40 & [1] \\ 1500y\cos x \:=\:10 & [2] \end{array}\) \(\displaystyle \text{Find }x\;(0^o \,\text{-}\, 90^o)\,\text{ and }y.\) Click to expand... \(\displaystyle \text{Divide [1] by [2]: }\;\frac{1500y\sin x}{1500y\cos x} \:=\:\frac{40}{10} \quad\Rightarrow\quad \tan x \:=\:4\) \(\displaystyle \text{Therefore: }\;x \:=\:\arctan(4) \;\approx\;\boxed{76^o}\) \(\displaystyle \text{Substitute into [1]: }\;1500y(\sin76^o) \:=\:40 \quad\Rightarrow\quad y \:=\:\frac{40}{1400\sin76^o}\) \(\displaystyle \text{Therefore: }\;y \:=\:0.027487371 \:\approx\:\boxed{0.027}\)
D Deleted member 4993 Guest Jun 23, 2009 #3 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)
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)
