Expand the right side of the equation using the compound angle formula for cosine.

The given equation is $7\cos 2x - 24\sin 2x = R\cos(2x+\alpha)$.
Using the formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$, we expand $R\cos(2x+\alpha)$:

$$R\cos(2x+\alpha) = R(\cos 2x \cos \alpha - \sin 2x \sin \alpha)$$ $$R\cos(2x+\alpha) = (R\cos\alpha)\cos 2x - (R\sin\alpha)\sin 2x$$

By comparing the expanded form with the left side of the given equation, $7\cos 2x - 24\sin 2x$:

1. $R\cos\alpha = 7$
2. $R\sin\alpha = 24$ (Note: The negative sign in front of $24\sin 2x$ matches the negative sign in the expanded formula, so $R\sin\alpha$ is positive.)

Square both equations from Step 2 and add them:

$$(R\cos\alpha)^2 + (R\sin\alpha)^2 = 7^2 + 24^2$$ $$R^2\cos^2\alpha + R^2\sin^2\alpha = 49 + 576$$ $$R^2(\cos^2\alpha + \sin^2\alpha) = 625$$

Since $\cos^2\alpha + \sin^2\alpha = 1$:

$$R^2 = 625$$

Given that $R > 0$:

$$R = \sqrt{625}$$ $$R = 25$$

Divide equation (2) by equation (1):

$$\frac{R\sin\alpha}{R\cos\alpha} = \frac{24}{7}$$ $$\tan\alpha = \frac{24}{7}$$

Given that $0 < \alpha < 90^\circ$, $\alpha$ is in the first quadrant.

$$\alpha = \arctan\left(\frac{24}{7}\right)$$

Using a calculator:

$$\alpha \approx 73.73979^\circ$$

Rounding to two decimal places:

$$\alpha \approx 73.74^\circ$$

The values are $R=25, \alpha \approx 73.74^\circ$.