Question

Find the slope of a line whose inclination to the positive direction of x-axis in anticlock wise direction is given as : (a) 15degree (b) 75 degree

Answer 1

$$\begin{gathered} \mathrm{Let}\mathit{}\theta \mathrm{be}\mathrm{the}\mathrm{inclination}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{line} \\ \mathrm{Now},\mathrm{slope},m=\tan \theta \\ \left(\mathbf{i}\right)\mathbf{.}\theta =15° \\ \mathrm{Therefore}m=\tan 15° \\ m=\tan \left(45°-30°\right) \\ =\frac{\tan 45°-\tan 30°}{1+\tan 45°\tan 30°} \\ =\frac{1-{\displaystyle \frac{1}{\sqrt{3}}}}{1+{\displaystyle \frac{1}{\sqrt{3}}}} \\ =\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ =\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)}\times \frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)} \\ =\frac{{\left(\sqrt{3}-1\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2} \\ =\frac{3+1-2\sqrt{3}}{3-1} \\ =\frac{4-2\sqrt{3}}{2} \\ =2-\sqrt{3} \end{gathered}$$

$$\begin{gathered} \left(\mathbf{ii}\right)\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathit{}\theta \mathbf{}\mathbf{=}75° \\ m=\tan 75° \\ =\tan \left(45°+30°\right) \\ =\frac{\tan 45°+\tan 30°}{1-\tan 45°\tan 30°} \\ =\frac{1+{\displaystyle \frac{1}{\sqrt{3}}}}{1-{\displaystyle \frac{1}{\sqrt{3}}}} \\ =\frac{\sqrt{3}+1}{\sqrt{3}-1} \\ =\frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)}\times \frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)} \\ =\frac{3+1+2\sqrt{3}}{3-1} \\ =\frac{4+2\sqrt{3}}{2} \\ =2+\sqrt{3} \end{gathered}$$