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**TwistedOne151** First, you rearrange $\displaystyle V_2\sin\theta-V_1d_1=d_2\cos\theta$ into

$\displaystyle V_2\sin\theta-d_2\cos\theta=V_1d_1$ via simple algebra.

Now, let $\displaystyle R=\sqrt{V_2^2+d_2^2}$, and let $\displaystyle \phi$ be an angle such that $\displaystyle tan\phi=\frac{d_2}{V_2}$ and such that $\displaystyle \sin\phi$ has the same sign as $\displaystyle d_2$. Then, we see that $\displaystyle V_2=R\cos\phi$ and $\displaystyle d_2=R\sin\phi$. Thus our equation becomes

$\displaystyle R\sin\theta\cos\phi-R\cos\theta\sin\phi=V_1d_1$

$\displaystyle \sin(\theta-\phi)=\frac{V_1d_1}{R}$

$\displaystyle \theta=\phi+\arcsin\frac{V_1d_1}{\sqrt{V_2^2+d_2^2 }}$

Note that if both $\displaystyle V_2$ and $\displaystyle d_2$ are non-negative, then $\displaystyle 0\le\phi\le\frac{\pi}{2}$, and $\displaystyle \phi=\arctan\left(\frac{d_2}{V_2}\right)$ with the usual range on the inverse tangent.

-Kevin C.