1. The problem asks to find the value of $\cos \theta$ in a right-angled triangle $MNP$ where $MN = \sqrt{7}$ cm and $MP = 4$ cm, with the right angle at $N$ and angle $\theta$ at $P$. 2. Recall the Pythagorean theorem for right triangles: $$NP^2 = MP^2 - MN^2$$ and the definition of cosine in a right triangle: $$\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}$$. 3. Calculate $NP$: $$NP^2 = 4^2 - (\sqrt{7})^2 = 16 - 7 = 9$$ $$NP = \sqrt{9} = 3$$ 4. Since $\theta$ is at $P$, the adjacent side to $\theta$ is $NP$ and the hypotenuse is $MP$. 5. Calculate $\cos \theta$: $$\cos \theta = \frac{NP}{MP} = \frac{3}{4} = 0.75$$ 6. Final answer: $$\boxed{\cos \theta = 0.75}$$