If 4cosA+3secA=4{3},0^{∘}

Correct Answer is : 30∘30^{∘}30∘

Correct Answer is : 30∘30^{∘}30∘

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SSC CHSL 15 Nov 2025 Shift 1 Paper

Question : 66 of 100

Marks: +1, -0

4cosA+3secA=4\sqrt{3},0^{∘}<^{′}A^{′}<90^{∘}

, then the value of ’A’ is:

Solution:

Concept:

Use the identity

\sec A = \frac{1}{\cos A}

to rewrite the equation in terms of

\cos A

, then solve the resulting quadratic equation.

Explanation:

Start with

4\cos A + 3\sec A = 4\sqrt{3}

Replace

\sec A

\frac{1}{\cos A}

4\cos A + \frac{3}{\cos A} = 4\sqrt{3}

Multiply both sides by

\cos A

(since

0^\circ < A < 90^\circ

\cos A > 0

4\cos^2 A + 3 = 4\sqrt{3}\cos A

Rearrange to form a quadratic:

4\cos^2 A - 4\sqrt{3}\cos A + 3 = 0

Let

x = \cos A

. Then

4x^2 - 4\sqrt{3}x + 3 = 0

Use the quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

with

a = 4

b = -4\sqrt{3}

c = 3

x = \frac{4\sqrt{3} \pm \sqrt{48 - 48}}{8} = \frac{4\sqrt{3} \pm 0}{8} = \frac{\sqrt{3}}{2}.

Thus

\cos A = \frac{\sqrt{3}}{2}

, which gives

A = 30^\circ

Answer:

30^\circ

(Option C).

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