S.2 Given: \( \cos \theta=\frac{-5}{7} \) and \( 0^{\circ}<\theta<180^{\circ} \). Determine \( \sin \left(\theta+60^{\circ}\right) \), without the use of a calculator.

To determine \( \sin \left(\theta + 60^{\circ}\right) \), we will use the sine addition formula: \[ \sin(\theta + 60^{\circ}) = \sin \theta \cos 60^{\circ} + \cos \theta \sin 60^{\circ} \] From trigonometric values, we know that \( \cos 60^{\circ} = \frac{1}{2} \) and \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \). Given \( \cos \theta = \frac{-5}{7} \), we need to find \( \sin \theta \). Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting \( \cos \theta \): \[ \sin^2 \theta + \left(\frac{-5}{7}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{25}{49} = 1 \] \[ \sin^2 \theta = 1 - \frac{25}{49} = \frac{49}{49} - \frac{25}{49} = \frac{24}{49} \] Thus, taking the square root gives us: \[ \sin \theta = \frac{\sqrt{24}}{7} = \frac{2\sqrt{6}}{7} \] Note that since \( \theta \) is in the second quadrant (where sine is positive), we don't consider the negative root. Now substituting \( \sin \theta \) and \( \cos \theta \) back into the sine addition formula: \[ \sin(\theta + 60^{\circ}) = \left(\frac{2\sqrt{6}}{7}\right)\left(\frac{1}{2}\right) + \left(\frac{-5}{7}\right)\left(\frac{\sqrt{3}}{2}\right) \] Calculating each term separately: \[ = \frac{2\sqrt{6}}{14} + \frac{-5\sqrt{3}}{14} \] \[ = \frac{2\sqrt{6} - 5\sqrt{3}}{14} \] Thus, the final result is: \[ \sin(\theta + 60^{\circ}) = \frac{2\sqrt{6} - 5\sqrt{3}}{14} \]