To solve the problem, we need to find the integer $p$ such that the product of sines of angles from $2^\circ$ to $90^\circ$ is equal to $\frac{p\sqrt{5}}{2^{50}}$.

### Method 1:
1. **Expression for $\sin(90x)$**:
$$\sin(90x) = \Im\left[ (\cos x + i \sin x)^{90} \right]$$
Using the binomial theorem, we expand:
$$(\cos x + i \sin x)^{90} = \sum_{k=0}^{90} \binom{90}{k} (\cos x)^{90-k} (i \sin x)^k$$
The imaginary part is:
$$\Im\left[ (\cos x + i \sin x)^{90} \right] = \sum_{n=0}^{45} (-1)^n \binom{90}{2n+1} \sin^{2n+1}(x) \cos^{90-2n-1}(x)$$

2. **Polynomial in $\sin(x)$**:
$$\frac{\sin(90x)}{\sin(x)\cos(x)} = P(\sin(x))$$
This is a polynomial in $\sin(x)$ of degree 88, with roots at $\sin(x) = \pm \sin(2^\circ), \pm \sin(4^\circ), \ldots, \pm \sin(88^\circ)$.

3. Constant and Leading Coefficient:
The constant term of $P(x)$ is 90, and the leading coefficient is:
$$\sum_{n=0}^{44} \binom{90}{2n+1} = \frac{(1+1)^{90} - (1-1)^{90}}{2} = 2^{89}$$

4. Product of Sines:
$$\frac{90}{2^{89}} = \prod_{n=-44, n \neq 0}^{44} \sin(2n) = (-1)^{44} \left(\prod_{n=1}^{44} \sin(2n)\right)^2$$
Thus:
$$\sin(90^\circ) \prod_{n=1}^{44} \sin(2n) = \sqrt{\frac{45}{2^{88}}} = \frac{3\sqrt{5}}{2^{44}}$$

5. **Finding $p$**:
$$p = 3 \cdot 2^6 = 192$$

### Method 2:
1. Root of Unity:
Let $\omega = e^{2\pi i/90}$. Then:
$$\prod_{n=1}^{45} \sin(2n^\circ) = \sum_{n=1}^{45} \frac{\omega^n - 1}{2i \omega^{n/2}}$$

2. Symmetry of Sine:
$$\prod_{n=1}^{45} \sin(2n^\circ) = \prod_{n=46}^{89} \sin(2n^\circ)$$

3. Square of Product:
$$\left| \prod_{n=1}^{45} \sin(2n^\circ) \right|^2 = \sum_{n=1}^{89} \frac{|\omega^n - 1|}{2} = \frac{90}{2^{89}}$$

4. Positive and Real Product:
$$\frac{\sqrt{45}}{2^{44}} = \frac{3\sqrt{5}}{2^{44}}$$

5. **Finding $p$**:
$$p = 3 \cdot 2^6 = 192$$

The final answer is $\boxed{192}$