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- Thread starter theBEAST
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ehild

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[tex]\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}[/tex]

[tex]\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}[/tex]

With these substitutions, you get a quadratic equation for tan(x/2).

ehild

- #3

vela

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$$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}} \sin\theta + \frac{B}{\sqrt{A^2+B^2}}\cos\theta\right)$$Now identify

\begin{align*}

\cos\phi &= \frac{A}{\sqrt{A^2+B^2}} \\

\sin\phi &= \frac{B}{\sqrt{A^2+B^2}}

\end{align*}then you have

$$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}(\cos\phi \sin\theta + \sin\phi\cos\theta) = \sqrt{A^2+B^2}\sin(\theta+\phi)$$where ##\tan\phi = B/A##.

In this particular problem, if you move the factor of 4 kN over to the righthand side, you have

$$3 \sin \theta - 0.45 \cos \theta = 2.5$$and ##\tan\phi = -0.45/3##, from which you should recognize ɸ as being the angle the line connecting A to the point where the force acts makes with the horizontal.

- #4

SammyS

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What system of equations.I tried squaring both sides but can't seem to simplify it. Any ideas on how this can be solved?

Thanks!

Edit: For those who are interested, I have attached the question that has prompted me to solve this system of equations.

I see none.

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ehild

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- #6

SammyS

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Thanks e .It is in the title. 12sinx-1.8cosx=10

ehild

Any important information in the title

Also, OP refers to "... this system of equations".

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