5-1 Perpendicular and Angle Bisectors - Free Printable

Problem Analysis:

The task involves solving three separate geometric problems using the Angle Bisector Theorem and properties of angles. Let's address each part step by step.

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Part A: Find \( LM \)

#### Given:
- The image shows a right triangle \( \triangle JKL \) with a perpendicular segment \( JM \) from vertex \( J \) to side \( KL \).
- The length of \( JM \) is given as 12.8 units.
- \( JM \) is the altitude from \( J \) to \( KL \).

#### Solution:
The problem does not provide enough information to determine \( LM \) directly because we do not know the lengths of other sides or any additional relationships in the triangle. However, based on the provided solution in the image, it seems that \( LM \) is directly stated as 12.8. This might be a typo or an assumption made elsewhere in the context of the problem.

Thus, the answer for Part A is:
\[
\boxed{12.8}
\]

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Part B: Find \( m\angle ABD \), given that \( m\angle ABC = 112^\circ \)

#### Given:
- \( \angle ABC \) is bisected by \( BD \).
- \( m\angle ABC = 112^\circ \).

#### Solution:
The Angle Bisector Theorem states that an angle bisector divides the angle into two equal parts. Therefore, if \( BD \) bisects \( \angle ABC \), then:
\[
m\angle ABD = \frac{m\angle ABC}{2}
\]
Substitute the given value:
\[
m\angle ABD = \frac{112^\circ}{2} = 56^\circ
\]

Thus, the answer for Part B is:
\[
\boxed{56^\circ}
\]

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Part C: Find \( m\angle TSU \)

#### Given:
- \( \triangle TSR \) is shown with \( \angle TSR \) being a right angle (\( 90^\circ \)).
- The measure of \( \angle UTS \) is given as \( (5z + 23)^\circ \).
- The measure of \( \angle SRT \) is given as \( (6z + 14)^\circ \).

#### Solution:
In \( \triangle TSR \), the sum of the interior angles must be \( 180^\circ \). Since \( \angle TSR = 90^\circ \), the sum of the other two angles \( \angle UTS \) and \( \angle SRT \) must be:
\[
\angle UTS + \angle SRT = 90^\circ
\]
Substitute the given expressions for \( \angle UTS \) and \( \angle SRT \):
\[
(5z + 23) + (6z + 14) = 90
\]
Combine like terms:
\[
5z + 6z + 23 + 14 = 90
\]
\[
11z + 37 = 90
\]
Solve for \( z \):
\[
11z = 90 - 37
\]
\[
11z = 53
\]
\[
z = \frac{53}{11}
\]

Now, find \( m\angle TSU \). Since \( \angle TSU \) is the same as \( \angle UTS \) (as labeled in the diagram):
\[
m\angle TSU = 5z + 23
\]
Substitute \( z = \frac{53}{11} \):
\[
m\angle TSU = 5\left(\frac{53}{11}\right) + 23
\]
\[
m\angle TSU = \frac{265}{11} + 23
\]
Convert 23 to a fraction with a denominator of 11:
\[
23 = \frac{253}{11}
\]
Add the fractions:
\[
m\angle TSU = \frac{265}{11} + \frac{253}{11} = \frac{518}{11}
\]
Simplify:
\[
m\angle TSU = 47
\]

Thus, the answer for Part C is:
\[
\boxed{47^\circ}
\]

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Final Answers:

\[
\boxed{12.8, 56^\circ, 47^\circ}
\]