Q201 / 270 · NDA 2026 (Phase 1) · Mathematics
Passage: The equation of a straight line which passes through the origin and makes an angle $\alpha$ $(0\leq\alpha\leq90^\circ)$ with the line $L:x+\sqrt{3}\,y+3\sqrt{3}=0$ is $x-\sqrt{3}\,y=0$.
Question: What is the value of $\alpha$ ?
Question: What is the value of $\alpha$ ?
- A$30^\circ$
- B$45^\circ$
- Correct answerC$60^\circ$
- D$90^\circ$
Solution
The slope of the line $L: x + \sqrt{3}\,y + 3\sqrt{3} = 0$ is $m_1 = -\dfrac{1}{\sqrt{3}}$.
The slope of the line $x - \sqrt{3}\,y = 0$ is $m_2 = \dfrac{1}{\sqrt{3}}$.
The angle $\alpha$ between the two lines is given by $\tan \alpha = \left| \dfrac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values of $m_1$ and $m_2$:
$\tan \alpha = \left| \dfrac{-\dfrac{1}{\sqrt{3}} - \dfrac{1}{\sqrt{3}}}{1 + \left(-\dfrac{1}{\sqrt{3}}\right)\left(\dfrac{1}{\sqrt{3}}\right)} \right|$
$\tan \alpha = \left| \dfrac{-\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}} \right| = \left| \dfrac{-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \right| = \sqrt{3}$
Since $0 \leq \alpha \leq 90^\circ$, we get $\alpha = 60^\circ$.
Answer: $60^\circ$
The slope of the line $x - \sqrt{3}\,y = 0$ is $m_2 = \dfrac{1}{\sqrt{3}}$.
The angle $\alpha$ between the two lines is given by $\tan \alpha = \left| \dfrac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values of $m_1$ and $m_2$:
$\tan \alpha = \left| \dfrac{-\dfrac{1}{\sqrt{3}} - \dfrac{1}{\sqrt{3}}}{1 + \left(-\dfrac{1}{\sqrt{3}}\right)\left(\dfrac{1}{\sqrt{3}}\right)} \right|$
$\tan \alpha = \left| \dfrac{-\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}} \right| = \left| \dfrac{-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \right| = \sqrt{3}$
Since $0 \leq \alpha \leq 90^\circ$, we get $\alpha = 60^\circ$.
Answer: $60^\circ$
Hence, the correct answer is option: 'C'.