MCQ
The angle between the line √3x – y – 2 = 0 and x – √3y + 1 = 0 is
- A15°
- ✓30°
- C45°
- D60°
Here, $m_1=\frac{-\sqrt{3}}{-1}=\sqrt{3}$,
$m_2=\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}$
Now, $\tan \theta=\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|$
$\tan \theta=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \frac{1}{\sqrt{3}}}\right|=\left|\frac{3-1}{2 \sqrt{3}}\right|=\frac{1}{\sqrt{3}}$
$\theta=30^{\circ}$
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