$\begin{align} & \text{Given that }5\cos \theta +12\sin \theta =R\cos (\theta -\alpha ),\text{ where }R\text{ and }\alpha \text{ are independent } \\ & \text{of }\theta \text{ and }R\text{ is positive, obtain the value of }R\text{ and Hence or otherwise, find the} \\ & \text{value of }\theta \text{ between }-{{180}^{\circ }}\,\text{and }{{180}^{\circ }}\text{ which satisfy the equation } \\ & 5\cos \theta +12\sin \theta =3\tfrac{1}{4} \\\end{align}$
$\begin{align} & 5\cos \theta +12\sin \theta =R\cos (\theta -\alpha ) \\ & 5\cos \theta +12\sin \theta =R(\cos \theta \cos \alpha +\sin \theta \sin \alpha ) \\ & \text{By comparing terms} \\ & 5\cos \theta =R\cos \theta \cos \alpha \\ & R\cos \alpha =5----(i) \\ & \text{sqaure both sides of }(i) \\ & {{R}^{2}}{{\cos }^{2}}\alpha =25----(ii) \\ & \text{Also } \\ & 12\sin \theta =R\sin \theta \sin \alpha \\ & 12=R\sin \alpha ----(iii) \\ & \text{Square both sides of (iii)} \\ & {{R}^{2}}{{\sin }^{2}}\alpha =144----(iv) \\ & \text{Add equation (ii) and (iv) together} \\ & {{R}^{2}}{{\sin }^{2}}\alpha +{{R}^{2}}\cos \alpha =144+25 \\ & {{R}^{2}}({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )=169 \\ & {{R}^{2}}=169 \\ & R=13 \\ & \text{Dividing (iii) by (i)} \\ & \frac{R\sin \alpha }{R\cos \alpha }=\frac{12}{5} \\ & \tan \alpha =2.5,\text{ }\alpha ={{\tan }^{-1}}2.5,\text{ }\alpha ={{89.9}^{\circ }} \\ & 5\cos \theta +12\sin \theta =13\cos (\theta -{{89.9}^{\circ }}) \\ & \text{Hence , solving }5\cos \theta +12\sin \theta =3\tfrac{1}{4} \\ & 13\cos (\theta -{{89.9}^{\circ }})=3.25 \\ & \cos (\theta -{{89.9}^{\circ }})=0.25 \\ & \theta -{{89.9}^{\circ }}={{75.31}^{\circ }} \\ & \theta ={{89.9}^{\circ }}+{{75.31}^{\circ }}={{164.40}^{\circ }} \\\end{align}$