KLM is a triangle with m(∠K)=30∘. If KL=10 units, and tan(M)=1/2, then find the length of KM.
Hint: Draw a picture and include the altitude from the angle L to the side KM.


Let's solve this step by step.

### Step 1: Analyze the given information
- We are given a triangle \( KLM \), with the measure of \( \angle K = 30^\circ \).
- \( KL = 10 \) units.
- \( \tan(\angle M) = \frac{1}{2} \).

We need to find the length of \( KM \).

### Step 2: Draw a diagram
Draw a triangle \( KLM \). Place \( K \) at the origin of a coordinate system, with \( L \) and \( M \) at two other points. Now, draw the altitude from point \( L \) perpendicular to \( KM \), meeting \( KM \) at point \( P \). This divides \( \triangle KLM \) into two right triangles, \( \triangle KLP \) and \( \triangle LPM \).

### Step 3: Use trigonometric relations
#### In \( \triangle KLP \):
- Since \( \angle K = 30^\circ \), and we know \( KL = 10 \), we can find the length of the altitude \( LP \) and the projection \( KP \) using basic trigonometry.
  
- For the altitude \( LP \), we use:
  \[
  \sin(30^\circ) = \frac{LP}{KL}
  \]
  \[
  \sin(30^\circ) = \frac{LP}{10}
  \]
  Since \( \sin(30^\circ) = \frac{1}{2} \):
  \[
  \frac{1}{2} = \frac{LP}{10} \quad \Rightarrow \quad LP = 5 \text{ units}.
  \]

- For the base \( KP \), we use:
  \[
  \cos(30^\circ) = \frac{KP}{KL}
  \]
  \[
  \cos(30^\circ) = \frac{KP}{10}
  \]
  Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
  \[
  \frac{\sqrt{3}}{2} = \frac{KP}{10} \quad \Rightarrow \quad KP = 5\sqrt{3} \text{ units}.
  \]

#### In \( \triangle LPM \):
- From the given \( \tan(\angle M) = \frac{1}{2} \), we know:
  \[
  \tan(\angle M) = \frac{LP}{PM} = \frac{5}{PM}.
  \]
  Therefore, \( PM = 10 \) units.

### Step 4: Find the length of \( KM \)
Now, we can find \( KM \) by adding the lengths of \( KP \) and \( PM \):
\[
KM = KP + PM = 5\sqrt{3} + 10.
\]
Thus, the length of \( KM \) is \( 5\sqrt{3} + 10 \) units.

### Summary
The length of \( KM \) is \( 5\sqrt{3} + 10 \) units.

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Would you like further clarification on any step, or do you have any questions?

Here are five follow-up questions to consider:
1. What is the geometric significance of the altitude in a triangle?
2. How can you use the tangent function to find other unknown sides in a right triangle?
3. How would the problem change if \( \tan(M) \) had a different value?
4. What would be the impact if \( \angle K \) were greater than \( 30^\circ \)?
5. Can the Pythagorean theorem be used here? If so, how?

**Tip:** In any triangle, altitudes are useful tools for creating right triangles and applying trigonometric functions effectively.

KLM is a triangle with m(∠K)=30∘. If KL=10 units, and tan(M)=1/2, then find the length of KM. Hint: Draw a picture and include the altitude from the angle L to the side KM.

Solve for the length of KM in triangle KLM, given that m(∠K)=30°, KL=10 units, and tan(M)=1/2. This problem involves using trigonometric identities, such as sine, cosine, and tangent functions, along with the Pythagorean theorem. Suitable for students in Grades 9-11.

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