Question

reduce the equation x+ root 3 y -4=0 to the normal form x cos alpha +y sin alpha = p and hence find the values of alpha and p

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}, \\ \mathrm{x}+\sqrt{3}\mathrm{y}-4=0 \\ \Rightarrow \mathrm{x}+\sqrt{3}\mathrm{y}=4 \\ \Rightarrow \frac{\mathrm{x}}{\sqrt{{\left(1\right)}^2+{\left(\sqrt{3}\right)}^2}}+\frac{\sqrt{3}\mathrm{y}}{\sqrt{{\left(1\right)}^2+{\left(\sqrt{3}\right)}^2}}=\frac{4}{\sqrt{{\left(1\right)}^2+{\left(\sqrt{3}\right)}^2}} \\ \Rightarrow \frac{\mathrm{x}}{2}+\frac{\sqrt{3}}{2}\mathrm{y}=\frac{4}{2} \\ \Rightarrow \frac{\mathrm{x}}{2}+\frac{\sqrt{3}}{2}\mathrm{y}=2 \\ \mathrm{So},\mathrm{we}\mathrm{get}\cos \mathrm{\alpha }=\frac{1}{2}\mathrm{and}\sin \mathrm{\alpha }=\frac{\sqrt{3}}{2}\mathrm{and}\mathrm{p}=2 \\ \mathrm{Since},\cos \mathrm{\alpha }>0\mathrm{and}\sin \mathrm{\alpha }>\hspace{0.17em}0,\mathrm{so}\mathrm{\alpha }\mathrm{lies}\mathrm{in}\mathrm{first}\mathrm{quadrant}. \\ \mathrm{Now},\frac{\sin \mathrm{\alpha }}{\cos \mathrm{\alpha }}=\frac{\sqrt{3}/2}{1/2} \\ \Rightarrow \tan \mathrm{\alpha }=\sqrt{3} \\ \Rightarrow \mathrm{\alpha }=60° \\ \mathrm{So},\mathrm{given}\mathrm{equation}\mathrm{in}\mathrm{normal}\mathrm{form}\mathrm{is}, \\ \boxed{\mathbf{x}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{60}\mathbf{°}\mathbf{}\mathbf{+}\mathbf{}\mathbf{y}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{60}\mathbf{°}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}} \end{gathered}$$