1. **State the problem:** We are given a triangle with sides 8 units and 7 units enclosing an angle $\theta$. The area of the triangle is $14\sqrt{3}$ square units. We need to find $\cos \theta$ as a fraction. 2. **Formula for the area of a triangle using two sides and the included angle:** $$\text{Area} = \frac{1}{2}ab\sin \theta$$ where $a=8$, $b=7$, and the area is $14\sqrt{3}$. 3. **Substitute the known values:** $$14\sqrt{3} = \frac{1}{2} \times 8 \times 7 \times \sin \theta$$ 4. **Simplify the right side:** $$14\sqrt{3} = 28 \sin \theta$$ 5. **Solve for $\sin \theta$:** $$\sin \theta = \frac{14\sqrt{3}}{28} = \frac{\sqrt{3}}{2}$$ 6. **Recall the Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ 7. **Substitute $\sin \theta = \frac{\sqrt{3}}{2}$:** $$\left(\frac{\sqrt{3}}{2}\right)^2 + \cos^2 \theta = 1$$ $$\frac{3}{4} + \cos^2 \theta = 1$$ 8. **Solve for $\cos^2 \theta$:** $$\cos^2 \theta = 1 - \frac{3}{4} = \frac{1}{4}$$ 9. **Take the square root:** $$\cos \theta = \pm \frac{1}{2}$$ 10. **Determine the correct sign:** Since $\sin \theta = \frac{\sqrt{3}}{2}$ corresponds to $\theta = 60^\circ$ or $120^\circ$, and the triangle sides suggest an acute angle, we take the positive value. **Final answer:** $$\cos \theta = \frac{1}{2}$$