To solve for the length of $LM$ in the given triangle, we'll use trigonometry. Here's the step-by-step process:

Given:

• $\angle LNM = 90^\circ$ (Right angle at $N$)
• $\angle KLM = 52^\circ 25'$ (Angle $KLM$)
• $\angle NLM = 25^\circ 30'$ (Angle $NLM$)
• $LN = 48$ units

Steps to Solve:

1. Calculate $\angle KML$:

   • Since $\angle KML$ is the exterior angle to $\triangle LNM$: $\angle KML = \angle KLM - \angle NLM = 52^\circ 25' - 25^\circ 30' = 26^\circ 55'$

2. Use the sine rule in $\triangle LNM$: Since $\angle LNM = 90^\circ$, we can use basic trigonometry: $\sin(\angle NLM) = \frac{LM}{LN}$ $\sin(25^\circ 30') = \frac{LM}{48}$

   To find $LM$: $LM = 48 \times \sin(25^\circ 30')$

3. Calculate $LM$: Use the sine value for $25^\circ 30'$: $\sin(25^\circ 30') \approx 0.4305$ $LM = 48 \times 0.4305 \approx 20.66 \text{ units}$

So, the length of $LM$ is approximately 20.66 units.

Would you like further details or have any questions about this solution?

Here are 5 related questions:

1. How do you calculate the area of $\triangle LNM$?
2. What is the length of $MN$ using the given angles and sides?
3. How do you determine $\angle LKN$ in the full triangle?
4. How would you apply the cosine rule in this triangle?
5. What would the problem look like if the given triangle was scalene?

Tip: In trigonometry, the sine rule is particularly useful when you know two angles and one side of a triangle, which allows you to find unknown sides.