SinA + 0.3CosA = 0.588 find angle A

wpgrady

New member
Joined
May 22, 2019
Messages
1
I have reduced a mechanical friction force problem down to SinA+0.3CosA = 0.588
how do I solve for A, I know that the answer is 17.6 degrees but how??
 
I have reduced a mechanical friction force problem down to SinA+0.3CosA = 0.588
how do I solve for A, I know that the answer is 17.6 degrees but how??
I'm probably making this too hard. (It's a bad habit.)

We know that \(\displaystyle sin(A) = \pm \sqrt{1 - cos^2(A)}\) where we take either the + or - depending on what quadrant we are in.

So we have two equations to solve.
\(\displaystyle + \sqrt{1 - ~ cos^2(A)} + 0.3~cos(A) = 0.588\)
and
\(\displaystyle - \sqrt{1 - ~ cos^2(A)} + 0.3~cos(A) = 0.588\)

They aren't as bad as they look. To start you off, let's take the top equation:
\(\displaystyle \sqrt{1 - ~ cos^2(A)} + 0.3~cos(A) = 0.588\)

\(\displaystyle \sqrt{1 - ~ cos^2(A)} = 0.588 - 0.3~cos(A)\)

\(\displaystyle \left ( \sqrt{1 - ~ cos^2(A)} \right ) ^2 = (0.588 - 0.3~cos(A) )^2\)

\(\displaystyle 1 - ~cos^2(A) = 0.345744 - 0.3528~cos(A) + 0.09~cos^2(A)\)

By making the substitution u = cos(A) this becomes a quadratic equation, which you can solve for u, then \(\displaystyle A = arccos(u)\)

See what you can do with this and let us know.

-Dan
 
Another approach is to make the equation look like sin(A+B), using the angle-sum identity for the sine (or cosine), by defining [MATH]B = \tan^{-1}(0.3)[/MATH]. I get the same result that way.
 
Top