In triangle $ABC$, given angle $A = 125.4^{\circ}$, side $b = 2.4$ cm, and side $c = 5$ cm, we need to find the length of side $a$.

1. Problem Description

In triangle $ABC$, given angle $A = 125.4^{\circ}$, side $b = 2.4$ cm, and side $c = 5$ cm, we need to find the length of side $a$.

2. Solution Steps

We can use the Law of Cosines to find the length of side $a$. The Law of Cosines states that:

$a^2 = b^2 + c^2 - 2bc \cos A$

Plugging in the given values, we have:

$a^2 = (2.4)^2 + (5)^2 - 2(2.4)(5) \cos(125.4^{\circ})$

$a^2 = 5.76 + 25 - 24 \cos(125.4^{\circ})$

Since $\cos(125.4^{\circ}) \approx -0.5798$,

$a^2 = 30.76 - 24(-0.5798)$

$a^2 = 30.76 + 13.9152$

$a^2 = 44.6752$

Taking the square root of both sides, we get:

$a = \sqrt{44.6752}$

$a \approx 6.684$

3. Final Answer

The length of side $a$ is approximately 6.684 cm.