In triangle $ABC$, we are given that angle $B = 135.5^{\circ}$, side $c = 8$ cm, and side $a = 5$ cm. We need to find the length of side $b$.

1. Problem Description

In triangle $ABC$, we are given that angle $B = 135.5^{\circ}$, side $c = 8$ cm, and side $a = 5$ cm. We need to find the length of side $b$.

2. Solution Steps

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

$b^2 = a^2 + c^2 - 2ac \cos(B)$

Plugging in the given values, we have:

$b^2 = 5^2 + 8^2 - 2(5)(8) \cos(135.5^{\circ})$

$b^2 = 25 + 64 - 80 \cos(135.5^{\circ})$

$b^2 = 89 - 80 \cos(135.5^{\circ})$

The cosine of $135.5^{\circ}$ is approximately $-0.7133$. Therefore,

$b^2 = 89 - 80(-0.7133)$

$b^2 = 89 + 57.064$

$b^2 = 146.064$

Taking the square root of both sides, we get:

$b = \sqrt{146.064}$

$b \approx 12.0857$

3. Final Answer

$b \approx 12.09$ cm (rounded to two decimal places)