\[ (r \cos \theta)^2 + (r \sin \theta)^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \] Or, \[ r^2 \cos^2 \theta + r^2 \sin^2 \theta = \frac{1}{4} + \frac{3}{4} \] Or, \[ r^2 (\cos^2 \theta + \sin^2 \theta) = \frac{1 + 3}{4} \] Or, \[ r^2 = 1 \] So, \[ r = \pm 1 \] Since \(0^\circ < \theta < 90^\circ\), therefore \(r = 1\) (Answer)