It's all about the angles inside a triangle. The speeds arranged into a triangle must be of the same shape as the triangle made up of target course, intercept course and target bearing. The size of the triangle, or the exact lengths of the sides in terms of real distance doesn't matter for getting there at the same time as he does. Because the time it takes is the same for all lengths of the triangle sides, it can be taken out of the equation. You reduce the problem into an 1 hour situation.
From that you can calculate the missing pieces of information inside the triangle. You have to make use of a math rule known as the law of sines: the ratio of side length to the sine of opposing angle, is the same for all side/corner combinations.
a/sin(alpha)=b/sin(beta)=c/sin(gamma)
If alpha is AOB (the corner occupied by the target), then a is own speed.
If b is target speed, then beta is lead-angle (deflection; the corner occupied by you)
C would then be the closing speed along the target bearing.
And gamma would be the angle of the intercept point corner. (track angle???)
You only have to know 3 (AOB, own speed, targetspeed) out of 4 to calculate the missing lead-angle b.
The angle of the intercept point corner/ trackangle is not really of interest in practical use, but it is required if you want to calculate the time it takes to get to the intercept point. Since the closing speed determines the time needed for the range to be reduced to zero. So, since the sum of all angles inside a triangle is 180, you can figure out the gamma angle. Plugging that into the: a/sin(alpha)=c/sin(gamma) equation you can now calculate c (closing speed). And ultimately the time it takes to the intercept point from range.
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