The given equations are $\sqrt{3}x+y$ = 1 ... (i) x + $\sqrt{3}y$ = 1... (ii) Since we have to find an angle between the two lines i.e., firstly we have to find the slopes of (i) and (ii). Slope of (i) = $m_1$ = - $\sqrt{3}$ ... (iii) Slope of (ii) = $m_2$ = $-\frac{1}{\sqrt{3}}$ ... (iv) Let θ be an angle between the given two lines, then tan θ = $\left|\frac{m_1-m_2}{1+m_1m_2}\right|$ = $\left|\frac{-\sqrt{3}-\left(-\frac{1}{\sqrt{3}}\right)}{1+\left(-\sqrt{3}\right)\cdot\left(-\frac{1}{\sqrt{3}}\right)}\right|$ (from (iii) & (iv)) = $\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+1}\right|$ = $\left|\frac{-3+1}{2\sqrt{3}}\right|$ = $\left|\frac{-1}{\sqrt{3}}\right|$ = $\frac{1}{\sqrt{3}}$ ⇒ tan θ = $\frac{1}{\sqrt{3}}$ ⇒ tan θ = tan 30° ⇒ θ = 30° ∴ The angle between the two given lines is 30° and the other angle is 180° – 30° = 150°.