1. Applying the Law of Cosines:
The given inequality is $ka^2 + lb^2 > c^2$. Using the Law of Cosines, we have:
$$c^2 = a^2 + b^2 - 2ab \cos{\angle{C}}$$
Substituting this into the inequality, we get:
$$ka^2 + lb^2 > a^2 + b^2 - 2ab \cos{\angle{C}}$$
Simplifying, we obtain:
$$(k-1)a^2 + (l-1)b^2 > -2ab \cos{\angle{C}}$$

2. **Considering the Range of $\cos{\angle{C}}$:**
Since $\cos{\angle{C}}$ can take values arbitrarily close to $-1$ (but not exactly $-1$ because $A$, $B$, and $C$ would be collinear), we must have:
$$(k-1)a^2 + (l-1)b^2 \geq 2ab$$

3. **Analyzing the Case When One of $k$ or $l$ is Not Larger Than 1:**
Suppose $k \leq 1$. Then:
$$(k-1)a^2 + (l-1)b^2 \leq (l-1)b^2$$
This implies:
$$(l-1)b^2 \geq 2ab$$
This inequality cannot hold for sufficiently large $a$. Therefore, both $k$ and $l$ must be greater than 1.

4. Ensuring the Inequality Holds:
We need to ensure:
$$(k-1)a^2 + (l-1)b^2 \geq 2ab \sqrt{(k-1)(l-1)}$$
If $(k-1)(l-1) \geq 1$, then:
$$(k-1)a^2 + (l-1)b^2 \geq 2ab$$
This condition is necessary for the initial inequality to hold.

5. Showing the Necessity of the Condition:
If $(k-1)(l-1) < 1$, then:
$$(k-1)a^2 - 2ab + (l-1)b^2 \geq 0$$
This is a quadratic inequality in $a$ with a non-positive discriminant:
$$D = (2b)^2 - 4(k-1)(l-1)b^2 = 4b^2(1 - (k-1)(l-1))$$
If $(k-1)(l-1) < 1$, then $D > 0$, which is a contradiction. Therefore, $(k-1)(l-1) \geq 1$ is necessary.

6. Conclusion:
The suitable pairs $(k, l)$ are those that satisfy:
$$(k-1)(l-1) \geq 1 \quad \text{and} \quad k, l > 1$$
This can be rewritten as:
$$kl \geq k + l \quad \text{and} \quad k, l > 1$$

The final answer is $\boxed{ (k, l) }$ such that $kl \geq k + l$ and $k, l > 1$.