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How to solve 12sinx-1.8cosx=10 without a graphing calculator?
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When you have a combination like ##A \sin\theta + B \cos\theta##, a useful trick is to write
$$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}} \sin\theta + \frac{B}{\sqrt{A^2+B^2}}\cos\theta\right)$$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.
$$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}} \sin\theta + \frac{B}{\sqrt{A^2+B^2}}\cos\theta\right)$$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.
