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:

• A. 50
• B. 40
• C. 52
• D. 54

Answer 1

Answer: (C)

52

Answer 2

The required area can be expressed as: $\text{Required Area} = \text{Area of Circle (from 0 to 2)} - \text{Area under Parabola (from 0 to 2)}.$ $\text{Required Area} = \int_0^2 \sqrt{8 - x^2} \, dx - \int_0^2 \sqrt{2x} \, dx$

We calculate the two integrals separately:

1. Area under the circle: $\int_0^2 \sqrt{8 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{8 - x^2} + \frac{8}{2} \sin^{-1} \frac{x}{\sqrt{8}} \right]_0^2.$

Substituting the limits: $\int_0^2 \sqrt{8 - x^2} \, dx = \frac{2}{2} \sqrt{8 - 4} + \frac{8}{2} \sin^{-1} \frac{2}{2\sqrt{2}} - \left( 0 + \frac{8}{2} \sin^{-1} 0 \right).$ $= 2 + 4 \cdot \frac{\pi}{4} = 2 + \pi.$

2. Area under the parabola: $\int_0^2 \sqrt{2x} \, dx = \left[ \frac{2}{3} (2x)^{3/2} \right]_0^2.$

Substituting the limits: $\int_0^2 \sqrt{2x} \, dx = \frac{2}{3} \cdot (2\sqrt{2}) - 0 = \frac{8}{3}.$

Thus, the required area is: $\text{Required Area} = (2 + \pi) - \frac{8}{3}.$

Simplifying: $\text{Required Area} = \pi - \frac{2}{3}.$