Find the length of LN using KNL.

In $\triangle KNL$, we are given:

• $\angle KNL = 90^\circ$ (indicated by the square symbol at N)
• $KL = 8.5$ cm (hypotenuse)
• $\angle K = 30^\circ$ We want to find the length of LN, which is the side opposite to $\angle K$. Using the sine function:

$$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$ $$\sin(30^\circ) = \frac{LN}{KL}$$

Substitute the given values:

$$\sin(30^\circ) = \frac{LN}{8.5 \text{ cm}}$$

We know that $\sin(30^\circ) = 0.5$:

$$0.5 = \frac{LN}{8.5 \text{ cm}}$$

Solve for LN:

$$LN = 0.5 \times 8.5 \text{ cm}$$ $$LN = 4.25 \text{ cm}$$

In $\triangle LNM$, we are given:

• $\angle NLM = 90^\circ$ (indicated by the square symbol at L)
• $LN = 4.25$ cm (calculated in Step 1)
• $\angle LNM = 55^\circ$ We want to find the length of MN, which is the hypotenuse of $\triangle LNM$. LN is the side adjacent to $\angle LNM$. Using the cosine function:

$$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\cos(\angle LNM) = \frac{LN}{MN}$$

Substitute the known values:

$$\cos(55^\circ) = \frac{4.25 \text{ cm}}{MN}$$

Solve for MN:

$$MN = \frac{4.25 \text{ cm}}{\cos(55^\circ)}$$

Calculate the value of $\cos(55^\circ)$:

$$\cos(55^\circ) \approx 0.573576$$

Now, calculate MN:

$$MN = \frac{4.25 \text{ cm}}{0.573576}$$ $$MN \approx 7.4099 \text{ cm}$$

$$MN \approx 7.4 \text{ cm}$$

The correct option is B.
$\boxed{\text{B. 7.4 cm}}$