## Solving: A sin (theta) + B cos (theta) + C = 0

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

### Solving: A sin (theta) + B cos (theta) + C = 0

From a physics questions with an airplane and wind, I get two equations (with angles in degrees) and must solve for angle A:

250 cos(A) = V cos(-23) - 80 cos(-45)
250 sin(A) = V sin(-23) - 80 sin(-45)

250 cos(A) = .92V - 56.6
250 sin(A) = -.39V + 56.6

Solving for V in the first equation...

V = (250 cos(A) + 56.6) / .92 = 271.6 cos(A) + 61.49

Substituting for V in the second equation...

250 sin(A) = -.39 (271.6 cos(A) + 61.49) +56.6

250 sin(A) = -106.1 cos(A) + 32.6

How do I solve for the angle A?

Thank you
cagey

Posts: 1
Joined: Mon Dec 07, 2009 5:43 am

cagey wrote:250 sin(A) = -106.1 cos(A) + 32.6

How do I solve for the angle A?

I think you could use an identity for linear combinations:

. . . . .$a\sin(\theta)\, +\, b\cos(\theta)\, =\, \sqrt{a^2\, +\, b^2}\,\sin(\theta\, +\, \psi)$

...where:

. . . . .$\psi\, =\, \arctan\left(\frac{b}{a}\right)\, +\, \left{\begin{array}{ll}0&\mbox{if}\, a\, \geq\, 0\\ \pi&\mbox{if}\, a\, <\, 0\end{array}$

This will give you an expression in just sine for the left-hand side, with 32.6 on the right-hand side. Divide through the square root, and then take the arcsine.

stapel_eliz

Posts: 1793
Joined: Mon Dec 08, 2008 4:22 pm