Tricky trigonometry question

[rachil0]

Heh, a pack of math nerds on the prowl Took 'er down like a limpy gazelle.

[Trugas]

Lmao, you guys are freaks!
Thankyou!! I've been struggling with this for days.

[Trugas]

Okay fun's not over yet though. I was just considering if this can be done by using the rule of sines too? Here's what I mean...

The dark gray triangle has two known sides and I think angle 'b' can be found too. So is that enough to find the third side of the triangle? If so then we can subtract that length from H and presto, blue triangle can be solved!

So Pythagora for totalDist:

totalDist = Math.sqrt((x + dx) * (x + dx) + (y + dy) * (y + dy)); 

and angle 'b' my trig is not so good but I'm pretty sure it could be done. Maybe find the top angle of the imaginary triangle just used in that last pythagora step above and subtract the top angle of the dotted line triangle (180 - (a + 90)).

[Trugas]

Would leave this to solve...

Question 1) Is this way even possible or am I assuming too much?
Question 2) Even if it does work would it just take longer to do it this way anyway?

[rachil0]

That might work, although it will have the same branch cut issue as the parametric solution. Recalling vaguely from geometry, SAS (side-angle-side) SSS and ASA all uniquely specify a triangle, but SSA does not. The figure you drew has this SSA arrangement and your triangle is not unique. I can make a second triangle with the same SSA - by picking the same other point, behind the car.

Not to say you couldn't tackle the problem going that way - always more than one way to get the job done. Wikipedia's article on the law of sines appears to have a special section on this, entitled "the ambiguous case".