To find the value of $\frac{3 \cos 36^{\circ}+5 \sin 18^{\circ}}{5 \cos 36^{\circ}-3 \sin 18^{\circ}}$ in the form $\frac{a \sqrt{5}-b}{c}$, we need to simplify the given expression. Let's start by using some fundamental trigonometric identities. We know that: $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$ $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$ First, substitute these values into the expression: $\frac{3 \cdot \frac{\sqrt{5} + 1}{4} + 5 \cdot \frac{\sqrt{5} - 1}{4}}{5 \cdot \frac{\sqrt{5} + 1}{4} - 3 \cdot \frac{\sqrt{5} - 1}{4}}$ Simplify the numerator and the denominator: Numerator: $3 \cdot \frac{\sqrt{5} + 1}{4} + 5 \cdot \frac{\sqrt{5} - 1}{4} = \frac{3(\sqrt{5} + 1) + 5(\sqrt{5} - 1)}{4} = \frac{3\sqrt{5} + 3 + 5\sqrt{5} - 5}{4} = \frac{8\sqrt{5} - 2}{4} = 2 \sqrt{5} - \frac{1}{2}$ Denominator: $5 \cdot \frac{\sqrt{5} + 1}{4} - 3 \cdot \frac{\sqrt{5} - 1}{4} = \frac{5(\sqrt{5} + 1) - 3(\sqrt{5} - 1)}{4} = \frac{5\sqrt{5} + 5 - 3\sqrt{5} + 3}{4} = \frac{2\sqrt{5} + 8}{4} = \frac{\sqrt{5}+4}{2}$ Now combine the simplified numerator and denominator: $\frac{2 \sqrt{5} - \frac{1}{2}}{\frac{\sqrt{5}+4}{2}} = \frac{(2 \sqrt{5} - \frac{1}{2}) \cdot 2}{\sqrt{5}+4} = \frac{4 \sqrt{5} - 1}{\sqrt{5}+4}$ Rationalizing the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator: $\frac{(4 \sqrt{5} - 1)(\sqrt{5}-4)}{(\sqrt{5}+4)(\sqrt{5}-4)}$ The denominator simplifies to: $\sqrt{5}^2 - 4^2 = 5 - 16 = -11$ The numerator simplifies to: $(4 \sqrt{5} - 1)(\sqrt{5}-4) = (4 \sqrt{5} \cdot \sqrt{5} - 4 \cdot 4 \sqrt{5} - 1 \cdot \sqrt{5} + 1 \cdot 4) = (20 - 16 \sqrt{5} - \sqrt{5} + 4) = 24 - 17 \sqrt{5}$ Combining them, we get: $\frac{24 - 17\sqrt{5}}{-11} = \frac{17\sqrt{5}-24}{11}$ Thus, $a = 17, b = 24, c = 11$. Therefore, $a + b + c = 17 + 24 + 11 = 52$. So the correct option is: Option B: 52