কোন রেখাটি 3y = √3x + 15 রেখায় 15^o কোণে অবণত?

Explanation:

Another Explanation (5): দেয়া আছে, \(3y = \sqrt{3}x + 15\) সুতরাং, \(y = \frac{\sqrt{3}}{3}x + 5\) \(y = \frac{1}{\sqrt{3}}x + 5\) এই সরলরেখার ঢাল, \(m_1 = \frac{1}{\sqrt{3}}\) ধরি, নির্ণেয় রেখার ঢাল \(m_2\)। \(15^\circ\) কোণে অবনত হওয়ার শর্তানুসারে, \[\tan 15^\circ = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\] আমরা জানি, \(\tan 15^\circ = 2 - \sqrt{3}\) \[2 - \sqrt{3} = \left| \frac{\frac{1}{\sqrt{3}} - m_2}{1 + \frac{1}{\sqrt{3}} m_2} \right|\] এখন, দুটি সম্ভাবনা: ১) \(\frac{\frac{1}{\sqrt{3}} - m_2}{1 + \frac{1}{\sqrt{3}} m_2} = 2 - \sqrt{3}\) \(\frac{1}{\sqrt{3}} - m_2 = (2 - \sqrt{3})(1 + \frac{m_2}{\sqrt{3}})\) \(\frac{1}{\sqrt{3}} - m_2 = 2 + \frac{2m_2}{\sqrt{3}} - \sqrt{3} - m_2\) \(\frac{1}{\sqrt{3}} - 2 + \sqrt{3} = \frac{2m_2}{\sqrt{3}}\) \(m_2 = \frac{\sqrt{3}}{2} (\frac{1}{\sqrt{3}} - 2 + \sqrt{3})\) \(m_2 = \frac{1}{2} - \sqrt{3} + \frac{3}{2} = 2 - \sqrt{3}\) ২) \(\frac{\frac{1}{\sqrt{3}} - m_2}{1 + \frac{1}{\sqrt{3}} m_2} = - (2 - \sqrt{3}) = \sqrt{3} - 2\) \(\frac{1}{\sqrt{3}} - m_2 = (\sqrt{3} - 2)(1 + \frac{m_2}{\sqrt{3}})\) \(\frac{1}{\sqrt{3}} - m_2 = \sqrt{3} + m_2 - 2 - \frac{2m_2}{\sqrt{3}}\) \(\frac{1}{\sqrt{3}} - \sqrt{3} + 2 = m_2 - \frac{2m_2}{\sqrt{3}} + m_2\) \(2 + \frac{1}{\sqrt{3}} - \sqrt{3} = 2m_2 - \frac{2m_2}{\sqrt{3}}\) \(2 + \frac{1 - 3}{\sqrt{3}} = m_2(2 - \frac{2}{\sqrt{3}})\) \(2 - \frac{2}{\sqrt{3}} = m_2(2 - \frac{2}{\sqrt{3}})\) \(m_2 = 1\) এখন, \(3y = 3x + 7\) রেখাটির ঢাল \(m = 1\)। সুতরাং, এই রেখাটি \(15^\circ\) কোণে অবনত। 🥳