SOLUTION: Geometry: Angle Relationships & Measures - Studypool - Free Printable

The image contains a worksheet titled "Using Angle Relationships" with several problems involving angles and algebra. Below, I will solve the problems step by step.

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Problem 1:

Given:
\( \angle A = 42^\circ \)
To find: The measure of \( \angle B \), where \( \angle A \) and \( \angle B \) are complementary angles.

Solution:
Complementary angles add up to \( 90^\circ \).
\[
\angle A + \angle B = 90^\circ
\]
Substitute \( \angle A = 42^\circ \):
\[
42^\circ + \angle B = 90^\circ
\]
Solve for \( \angle B \):
\[
\angle B = 90^\circ - 42^\circ = 48^\circ
\]

Answer:
\[
\boxed{48^\circ}
\]

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Problem 2:

Given:
\( \angle 1 = 113^\circ \)
To find: The measure of \( \angle 2 \), where \( \angle 1 \) and \( \angle 2 \) form a linear pair.

Solution:
Linear pairs of angles add up to \( 180^\circ \).
\[
\angle 1 + \angle 2 = 180^\circ
\]
Substitute \( \angle 1 = 113^\circ \):
\[
113^\circ + \angle 2 = 180^\circ
\]
Solve for \( \angle 2 \):
\[
\angle 2 = 180^\circ - 113^\circ = 67^\circ
\]

Answer:
\[
\boxed{67^\circ}
\]

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Problem 3:

Given:
\( \angle SQR = (x + 8)^\circ \)
\( \angle TQS = 47^\circ \)
To find: The measure of \( \angle SQR \).

Solution:
Since \( \angle SQR \) and \( \angle TQS \) are adjacent angles forming a straight line, their sum is \( 180^\circ \).
\[
\angle SQR + \angle TQS = 180^\circ
\]
Substitute \( \angle SQR = (x + 8)^\circ \) and \( \angle TQS = 47^\circ \):
\[
(x + 8)^\circ + 47^\circ = 180^\circ
\]
Solve for \( x \):
\[
x + 8 + 47 = 180
\]
\[
x + 55 = 180
\]
\[
x = 180 - 55 = 125
\]
Now, substitute \( x = 125 \) back into \( \angle SQR = (x + 8)^\circ \):
\[
\angle SQR = (125 + 8)^\circ = 133^\circ
\]

Answer:
\[
\boxed{133^\circ}
\]

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Problem 4:

Given:
\( \angle DCE = (7x - 25)^\circ \)
\( \angle ECF = (4x + 19)^\circ \)
To find: The measure of \( \angle DCE \).

Solution:
Since \( \angle DCE \) and \( \angle ECF \) are vertical angles, they are equal.
\[
\angle DCE = \angle ECF
\]
Substitute the given expressions:
\[
7x - 25 = 4x + 19
\]
Solve for \( x \):
\[
7x - 4x = 19 + 25
\]
\[
3x = 44
\]
\[
x = \frac{44}{3}
\]
Now, substitute \( x = \frac{44}{3} \) back into \( \angle DCE = (7x - 25)^\circ \):
\[
\angle DCE = 7\left(\frac{44}{3}\right) - 25
\]
\[
\angle DCE = \frac{308}{3} - 25
\]
Convert 25 to a fraction with a denominator of 3:
\[
25 = \frac{75}{3}
\]
\[
\angle DCE = \frac{308}{3} - \frac{75}{3} = \frac{233}{3}
\]

Answer:
\[
\boxed{\frac{233}{3}}
\]

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Problem 5:

Given:
\( \angle KNM = (8x - 5)^\circ \)
\( \angle LMN = (7x + 7)^\circ \)
\( \angle KNL = 124^\circ \)
To find: The measure of \( \angle KNM \).

Solution:
The angles around point \( N \) form a full circle, so their sum is \( 360^\circ \).
\[
\angle KNM + \angle LMN + \angle KNL = 360^\circ
\]
Substitute the given expressions:
\[
(8x - 5)^\circ + (7x + 7)^\circ + 124^\circ = 360^\circ
\]
Combine like terms:
\[
8x + 7x - 5 + 7 + 124 = 360
\]
\[
15x + 126 = 360
\]
Solve for \( x \):
\[
15x = 360 - 126
\]
\[
15x = 234
\]
\[
x = \frac{234}{15} = 15.6
\]
Now, substitute \( x = 15.6 \) back into \( \angle KNM = (8x - 5)^\circ \):
\[
\angle KNM = 8(15.6) - 5
\]
\[
\angle KNM = 124.8 - 5 = 119.8
\]

Answer:
\[
\boxed{119.8^\circ}
\]

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Final Answers:

1. \( \boxed{48^\circ} \)
2. \( \boxed{67^\circ} \)
3. \( \boxed{133^\circ} \)
4. \( \boxed{\frac{233}{3}} \)
5. \( \boxed{119.8^\circ} \)