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UPSC NDA I 2026 Mathematics Paper

Question Numbers: 51-52

For the next two (02) items that follow:

The equation of a straight line which passes through the origin and makes an angle

\alpha\left(0 \leqslant \alpha \leqslant 90^{\circ}\right)

with the line

\mathrm{L}: \mathrm{x}+\sqrt{3} y+3 \sqrt{3}=0

\mathrm{x}-\sqrt{3} y=0

Question : 51 of 120

Marks: +1, -0

What is the value of

\alpha

Solution:

Concept:

Angle between two lines using their slopes:

\tan \alpha = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Explanation:

Line

L

x + \sqrt{3}y + 3\sqrt{3} = 0

gives

y = -\frac{1}{\sqrt{3}}x - 3

So,

m_1 = -\frac{1}{\sqrt{3}}

Given line:

x - \sqrt{3}y = 0

gives

y = \frac{1}{\sqrt{3}}x

So,

m_2 = \frac{1}{\sqrt{3}}

Now,

\tan \alpha = \left| \frac{\frac{1}{\sqrt{3}} - \left(-\frac{1}{\sqrt{3}}\right)}{1 + \left(-\frac{1}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right)} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} \right| = \sqrt{3}

Therefore,

\alpha = 60^\circ

Answer:

60^\circ

, i.e. Option C.

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