1. The problem asks us to find two consecutive whole numbers between which the value of $\sqrt{75}$ lies. 2. Recall that $\sqrt{n}$ is the number which, when squared, equals $n$. 3. We need to find whole numbers $a$ and $b$ such that: $$a < \sqrt{75} < b$$ where $a$ and $b$ are consecutive whole numbers. 4. First, find perfect squares near 75: $$8^2 = 64 \quad \text{and} \quad 9^2 = 81$$ 5. Since $64 < 75 < 81$, it follows that: $$8^2 < 75 < 9^2$$ 6. Taking square roots (which preserves inequalities for positive numbers): $$8 < \sqrt{75} < 9$$ 7. Therefore, $\sqrt{75}$ lies between the consecutive whole numbers 8 and 9. Final answer: $$8 < \sqrt{75} < 9$$