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

θ

(in degrees):

$950 \sin θ - 950 \tan 15 \cos θ = 175 \sin 25 - 175 \tan 15 \cos 25$

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

But how? Trying to do it without the ''solver'' I can get as far as $C_{1} \sin θ - C_{2} \cos θ = C_{3}$ . How would you solve that for $θ$ ?

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 (θ - φ) = \cos φ \sin θ - \sin φ \cos θ

Can you see how to find $A$ and $φ$ such that $C_{1} = A \cos φ$ , $C_{2} = A \sin φ$ ? Then we can rewrite the equation as $\sin (θ - φ) = C_{3} / A$ , 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.

φ = \arctan (C_{2} / C_{1})

. Then solve for

A

.

Great trick! Thanks, David.