Question:
If the value of \[ \frac{3 \cos 36^\circ + 5 \sin 18^\circ}{5 \cos 36^\circ - 3 \sin 18^\circ} = \frac{a\sqrt{5} - b}{c}, \] where \(a, b, c\) are natural numbers and \(\text{gcd}(a, c) = 1\), then \(a + b + c\) is equal to:
If the value of \[ \frac{3 \cos 36^\circ + 5 \sin 18^\circ}{5 \cos 36^\circ - 3 \sin 18^\circ} = \frac{a\sqrt{5} - b}{c}, \] where \(a, b, c\) are natural numbers and \(\text{gcd}(a, c) = 1\), then \(a + b + c\) is equal to:
Updated On: Mar 19, 2026
- 50
- 40
- 52
- 54
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
\( \dfrac{3\left(\dfrac{\sqrt{5}+1}{4}\right) + 5\left(\dfrac{\sqrt{5}-1}{4}\right)}{5\left(\dfrac{\sqrt{5}+1}{4}\right) - 3\left(\dfrac{\sqrt{5}-1}{4}\right)} = \dfrac{8\sqrt{5}-2}{2\sqrt{5}+8} \)
\( = \dfrac{4\sqrt{5}-1}{\sqrt{5}+4} \times \dfrac{\sqrt{5}-4}{\sqrt{5}-4} \)
\( = \dfrac{20 - 16\sqrt{5} - \sqrt{5} + 4}{-11} \)
\( = \dfrac{17\sqrt{5}-24}{11} \Rightarrow a = 17, \, b = 27, \, c = 11 \)
\( a + b + c = 52 \)
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Trigonometry Questions
- If \( 2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0 \) has exactly 3 solutions in the interval \( \left[ 0, \frac{n \pi}{2} \right] \), \( n \in \mathbb{N} \), then the roots of the equation \( x^2 + nx + (n - 3) = 0 \) belong to:
- \[\text{Let } A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix} \text{ and } P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}.\]The sum of the prime factors of \( |P^{-1}AP - 2I| \) is equal to.
- Let \( x = \frac{m}{n} \) ( \( m, n \) are co-prime natural numbers) be a solution of the equation \( \cos \left( 2 \sin^{-1} x \right) = \frac{1}{9} \) and let \( \alpha, \beta (\alpha > \beta) \) be the roots of the equation \( mx^2 - nx - m + n = 0 \). Then the point \( (\alpha, \beta) \) lies on the line
- For \( \alpha, \beta \in \left( 0, \frac{\pi}{2} \right) \), let \( 3 \sin(\alpha + \beta) = 2 \sin(\alpha - \beta) \) and a real number \( k \) be such that \( \tan \alpha = k \tan \beta \). Then the value of \( k \) is equal to:
- Let the set of all \( a \in \mathbb{R} \) such that the equation \(\cos 2x + a \sin x = 2a - 7\) has a solution be \([p, q]\) and \( r = \tan 9^\circ - \tan 27^\circ - \frac{1}{\cot 63^\circ + \tan 81^\circ} \), then \( pqr \) is equal to ______.
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions