Joel Kammet
Posted on Sunday, 23 February, 2003 - 07:09 pm:
While working out a 2-dimensional motion problem (plane flying one way, wind blowing another, etc) I came up with this equation to solve for q (in degrees):

950 sinq-950 tan 15 cosq = 175 sin 25 - 175 tan 15 cos 25

The ''solver'' function of my calculator found q = 16.833 degrees.

But how? Trying to do it without the ''solver'' I can get as far as C1 sinq-C2 cosq = C3. How would you solve that for q?



BTW I'm convinced that the solution of 16.833 degrees is correct, because I rotated the axes by 15 deg. & was then able to solve for the x and y components separately, and came up with the same result (after re-adjusting the axes).

Much simpler to do it that way, but I'm still wondering how the calculator did it.

David Loeffler
Posted on Sunday, 23 February, 2003 - 07:23 pm:
Use the identity sin(q-j) = cosjsinq-sinjcosq.

Can you see how to find A and j such that C1=A cosj, C2=A sinj? Then we can rewrite the equation as sin(q-j)=C3/A, and you can immediately write down the solution to that.

(Incidentally, your calculator almost certainly used a numerical approach such as interval bisection or Newton's method, rather than this trick.)

David

Joel Kammet
Posted on Sunday, 23 February, 2003 - 07:52 pm:
Ah. j = arctan(C2/C1). Then solve for A.

Great trick! Thanks, David.