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Drop a height $ h$ from $ I$ down to side $ MT$. Then

$\displaystyle \sin \theta$ $\displaystyle = \frac{ h }{ r_{TI}\Delta t }$    
$\displaystyle \sin \alpha$ $\displaystyle = \frac{ h }{ r_{MI}\Delta t }$    

and so

$\displaystyle \sin \theta \cdot r_{TI}\Delta t = \sin \alpha \cdot r_{MI}\Delta t
\qquad \textrm{(Law of Sines)}

Through time the angles of the representation triangle are constant and the sides are proportional by some factor $ \Delta t_{2}/\Delta
t_{1}$: when solving for the angles, time may be eliminated from both the equation and the representation of the problem.

$\displaystyle \sin \theta \cdot r_{TI} = \sin \alpha \cdot r_{MI}

Since point $ I$ is not moving and thus $ v_{T} = r_{TI}$ and $ v_{M} =
r_{MI}$, $ \alpha $ can be expressed by the equation

$\displaystyle \sin \alpha = \sin \theta \cdot \frac{ v_{T} }{ v_{M} }

Mathematically, the solution for $ \alpha $ is ambiguous, since $ \sin
\alpha = \sin (180 - \alpha)$. However, in the physical situation it is clear that the $ \alpha $ desired is always acute; otherwise $ M$ would try and catch $ T$ by running away from it rather than towards it--a sometimes successful but always inefficient strategy. (The two solutions for $ \alpha $ are pictured in Figure 3; $ \alpha_{1}$ is always preferable to $ \alpha_{2}$, because a smaller $ \alpha $ means $ M$ is heading more directly towards the current location of $ T$. Alternately: according the representation, a smaller $ \alpha $ maximizes the length of $ MT$, which represents $ r_{MT}$, the closure rate between points $ M$ and $ T$.)

Figure 3: Ambiguous interception (mathematical representation).
Image ambiguous-interception
Assuming the implementation of the $ \arcsin$ function returns only the acute or right angle, the best solution for $ \alpha $ can be simply stated

$\displaystyle \alpha = \arcsin \left( \sin \theta \cdot \frac{ v_{T} }{ v_{M} } \right)$ (1)

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