Solving for a

**JohnC** · June 5th 2008, 03:14 PM

Original Equation:

$\sqrt{\frac{5 \cdot G \cdot D}{M^2 \cdot sin(2 \cdot A)}} \cdot 4 = P$

Solving for D:

$\frac{M^2 \cdot P^2 \cdot sin(2 \cdot A)} { 80 \cdot G } = D$

Solving for A (Don't think is right):

$\frac {sin^{-1} \left (\frac{80 \cdot G \cdot D} { M^2 \cdot P^2} \right )} {2} = A<br />$

I changed my original post after Dan replied to it. The equation was displayed wrong. I am trying to solve for A for one equation, and D for another. This is what I got after seeing Dan's work on how to work with sine. I would love to be double checked.

**topsquark** · June 5th 2008, 03:38 PM

Originally Posted by JohnC

Original Equation
.._____________
√98*(800(D/40)) * 4 = P
4*(240²)*SIN(2A)

or

4*SQRT((98 (800 (D/40))/(4(240²)sin(2A))) = P

------------------------

98(800(D/40)) = sin(2A)
(P/4)²*4(240)²

------------------------

That's about as far I got. I forgot how to get rid of sine since I haven't done it in about 2 years. If some one can tell me how I can go about getting rid of the sine, or alternatively, solving the problem for A, that would be amazing. I kept the numbers unsimplified for a reason. If you wish to solve for it and simplify it, 98 = G, and 240 = M. Otherwise, I would love to know how to get rid of the sin and solve for A.

You have to use the inverse of the sine function: $sin^{-1}$ or "asn" (arcsine.) Also, you didn't need to square both sides to get rid of the square root.
$\frac{\sqrt{98*(800(D/40))}}{4*(240²)*sin(2A)} \cdot 4 = P$

$sin(2A) = \frac{\sqrt{98*(800(D/40))}}{4*(240²)*P} \cdot 4$

Now take the inverse sine:
$2A = sin^{-1} \left ( \frac{\sqrt{98*(800(D/40))}}{4*(240²)*P} \cdot 4 \right )$

$A = \frac{1}{2} \cdot sin^{-1} \left ( \frac{\sqrt{98*(800(D/40))}}{4*(240²)*P} \cdot 4 \right )$

-Dan

**JohnC** · June 5th 2008, 08:03 PM

I updated my main post so the equation is easier to read. Thanks for reminding me to use arcsine

**Reckoner** · June 5th 2008, 11:16 PM

Originally Posted by JohnC

Original Equation:

$\sqrt{\frac{5 \cdot G \cdot D}{M^2 \cdot sin(2 \cdot A)}} \cdot 4 = P$

Solving for D:

$\frac{M^2 \cdot P^2 \cdot sin(2 \cdot A)} { 80 \cdot G } = D$

Solving for A (Don't think is right):

$\frac {sin^{-1} \left (\frac{80 \cdot G \cdot D} { M^2 \cdot P^2} \right )} {2} = A<br />$

Are there any restrictions on $G,\;D,\;M,\;\text{or}\;A$ ? As it is, we have

$4\sqrt{\frac{5\,GD}{M^2\sin(2A)}} = P$

$\Rightarrow16\left\lvert\frac{5\,GD}{M^2\sin(2A)}\ right\rvert = P^2$

$\Rightarrow\frac{80|GD|}{M^2\left\lvert\sin(2A)\ri ght\rvert} = P^2$

$\Rightarrow|D| = \frac{M^2P^2\left\lvert\sin(2A)\right\rvert}{80|G| }$

$\Rightarrow D = \pm\frac{M^2P^2\left\lvert\sin(2A)\right\rvert}{80 |G|}$

and, for $A$ ,

$|\sin 2A| = \frac{80|GD|}{M^2P^2}$

$\sin 2A = \pm\frac{80|GD|}{M^2P^2}$

$\Rightarrow A = \frac12\left[\arcsin\left(\pm\frac{80|GD|}{M^2P^2}\right) + 2n\pi\right],\;n\in\mathbb{Z}$ or $A = \frac12\left[(2n + 1)\pi - \arcsin\left(\pm\frac{80|GD|}{M^2P^2}\right)\right],\;n\in\mathbb{Z}$

This can all possibly be simplified quite a bit if more information is given about the variables involved.

**JohnC** · June 6th 2008, 08:45 AM

For the problem I am currently working on:
A = Angle, (0° - 90°)
D = Distance, (0 - 120)
G = Constant, 98
M = Constant, 240
P = Power, (0 - 4)

On a side note, and I think this will be too complicated for me, but:
How would one go about calculating the equation of an arc given just an image. I would guess Trial and Error to match the image?

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