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# V=(2.21x)/cos(a)*sqrt(-y+tan(a)*x) Get a by itself

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V=(2.21x)/cos(a)*sqrt(-y+tan(a)*x) Get a by itself

Guest Apr 12, 2015

#2
+889
+5

I don't think that you're going to get very far with this, unless there happens to be some nice relationship between x and y.

The way to deal with it is to take the cos(a) into the sq root as sec^2(a) and to replace it with 1+tan^2(a). That creates a cubic in tan(a) which will remain, even when the square root is removed.

Bertie  Apr 12, 2015
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#1
+92254
+5

$$\\V=\frac{(2.21x)}{cos(a)}*\sqrt{-y+tan(a)*x}\\ V^2=\frac{(2.21x)^2}{cos^2(a)}*(-y+tan(a)*x)\\ V^2cos^2a=(2.21x)^2*(-y+tan(a)*x)\\ V^2cos^2a=-(2.21x)^2y+(2.21x)^2tan(a)*x\\ V^2cos^2a+(2.21x)^2y=(2.21x)^2tan(a)*x\\ V^2cos^2a+(2.21x)^2y=(2.21x)^2\sqrt{\frac{1}{cos^2a}-1}*x\\ V^4cos^4a+2*V^2cos^2a*(2.21x)^2y+(2.21x)^4y^2=(2.21x)^4*(\frac{1}{cos^2a}-1)}*x^2\\ V^4cos^4a+2*V^2cos^2a*(2.21x)^2y+(2.21x)^4y^2=(2.21x)^4*(\frac{1-cos^2a}{cos^2a})*x^2\\ V^4cos^6a+2*V^2cos^4a*(2.21x)^2y+(2.21x)^4y^2cos^2a=(2.21x)^4*({1-cos^2a})*x^2\\$$

I am not getting anywhere with this.  Sorry.

Melody  Apr 12, 2015
#2
+889
+5

I don't think that you're going to get very far with this, unless there happens to be some nice relationship between x and y.

The way to deal with it is to take the cos(a) into the sq root as sec^2(a) and to replace it with 1+tan^2(a). That creates a cubic in tan(a) which will remain, even when the square root is removed.

Bertie  Apr 12, 2015
#3
+92254
0

Thanks Bertie

Melody  Apr 13, 2015

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