Solve 33cos(theta) - 46.2sin(theta) = 53.95 for theta

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mcwang719

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I wasn't sure if this goes under algebra or trig. I've been working on this problem for about half an hour now and I think I'm missing something.

. . .Solve the following for theta:

. . .33cos(theta) - 46.2sin(theta) = 53.95

I was tried to do was to get sin/cos to get inverse tan. Can anyone help? Thank you!
 
Is this the complete statement of the exercise, or is this a step in some other exercise? If the latter, please provide the original exercise, showing how you got down to this equation.

Thank you.

Eliz.
 
This problem is from my statics class, my question is a one little step that I got stuck on, there's a lot to this problem. I drew the freebody diagram, sum the forces in the x and in y and basically, what I need to do is find the range of values of (theta). I have the solutions and in the solutions the answer is theta=-36.3 and -72.6. thanks!
 
Hello, mcwang719!

You could work on this for days (weeks?) and never figure it out.
There is a special "trick" for this type of equation:\(\displaystyle \;a\,\cos\theta\,\pm\,b \,\sin \theta\;=\;c\)

First, find \(\displaystyle d\:=\:\sqrt{a^2\,+\,b^2}\)

Divide through by \(\displaystyle d:\;\) \(\displaystyle \L\;\;\underbrace{\left(\frac{a}{\sqrt{a^2+b^2}}\right)}\cos\theta\,\pm\,\underbrace{\left(\frac{b}{\sqrt{a^2+b^2}}\right)}\sin\theta\;= \;\frac{c}{\sqrt{a^2+b^2}}\)
Then we substitute: . . . . . \(\displaystyle \sin\alpha\) . . . . . . . . . . . .\(\displaystyle \cos\alpha\)

So we have: \(\displaystyle \L\,\sin\alpha\cos\theta\,\pm\,\cos\alpha\sin\theta \;= \;\frac{c}{\sqrt{a^2+b^2}}\)

And with: \(\displaystyle \L\,\sin(\alpha\,\pm\,\theta) \;= \;\frac{c}{\sqrt{a^2+b^2}}\) we can solve for \(\displaystyle \theta\).

Solve the following for \(\displaystyle \theta:\;\;33\,\cos\theta \,- \,46.2\,\sin\theta \:=\:53.95\)
Click to expand...

We have: \(\displaystyle \,d\:=\:\sqrt{33^2\,+\,46.2^2} \:\approx\:56.775\)

Divide by \(\displaystyle d:\) \(\displaystyle \L\:\frac{33}{56.775}\cos\theta\,-\,\frac{46.2}{56.775}\sin\theta\;=\;\frac{53.95}{56.775}\)

Then: \(\displaystyle \,\sin\alpha \:= \:\frac{33}{56.775}\;\;\Rightarrow\;\;\alpha\:\approx\:35.537^o\)

(Check: \(\displaystyle \cos\alpha\:=\:\frac{46.2}{56.775}\;\;\Rightarrow\;\;\alpha\:\approx\:35.537^o\))


We have: \(\displaystyle \,\sin35.537^o\cos\theta\,-\,\cos35.537^o\sin\theta\:=\:0.9502\)

Then: \(\displaystyle \,\sin(35.537^o\,-\,\theta)\;=\;0.9502\)

And: \(\displaystyle \,35.537^o\,-\,\theta\;=\;\sin^{-1}(0.9502)\:\approx\:71.85^o\) or \(\displaystyle 108.15^o\)


We have two equations to solve:

\(\displaystyle \;\;35.537^o\,-\,\theta\:=\:72.85^o\;\;\Rightarrow\;\;\theta\:=\;-36.313^o\)

\(\displaystyle \;\;35.537^o\,-\,\theta\:=\:108.15^o\;\;\Rightarrow\;\;\theta\:=\:-72.613^o\)

\(\displaystyle \;\;\) . . . ta-DAA!
 
Well, if I can't get any help here I'll see the professor. I was able to get break it down to 33=-46.25tan(theta), (don't know if it's right) so I take the inverse tan to get 35.51 degrees which is not the correct answer, and I don't see how they come up with two values for theta? thanks.
 
Wow! that's a tricky one, your right without your help I don't think I would have been able to solve that on my own. Thanks a lot soraban!
 
Soroban, I like that cool trick. Nice. I am going to put that one in my file.

Anyway, a rather boring method is ol' Newton.

Derivative of \(\displaystyle \L\\33cos({\theta})-\frac{231}{5}sin({\theta})=\frac{1079}{20}\)

is \(\displaystyle \L\\\frac{-231}{5}cos({\theta})-33sin({\theta})\)

Use Newton method formula:

\(\displaystyle \L\\{\theta}_{n+1}={\theta}-\frac{33cos({\theta})-\frac{231}{5}sin({\theta})-\frac{1079}{20}}{\frac{-231}{5}cos({\theta})-33sin({\theta})}\)

Start with an inital guess of 1 or 1.5.

It only takes a few iterations to arrive at acceptable accuracy.

I got \(\displaystyle {\theta}=-1.26734937329\;\ and \;\ -.633744308333\)

This is radians. In degrees they translate to -36.31 and -72.613
 
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