SOLUTION: Angle Addition Postulate Geometry Basics Worksheet ... - Free Printable

Let's solve each problem step by step.

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Problem 6

Given:
- \( m\angle CDF = (3x + 14)^\circ \)
- \( m\angle FDE = (5x - 2)^\circ \)
- \( m\angle CDE = (10x - 18)^\circ \)

We know that \( \angle CDE \) is the sum of \( \angle CDF \) and \( \angle FDE \):
\[
m\angle CDE = m\angle CDF + m\angle FDE
\]
Substitute the given expressions:
\[
10x - 18 = (3x + 14) + (5x - 2)
\]
Simplify the right-hand side:
\[
10x - 18 = 3x + 5x + 14 - 2
\]
\[
10x - 18 = 8x + 12
\]
Isolate \( x \):
\[
10x - 8x = 12 + 18
\]
\[
2x = 30
\]
\[
x = 15
\]

Now, substitute \( x = 15 \) back into the expressions for each angle:
\[
m\angle CDF = 3x + 14 = 3(15) + 14 = 45 + 14 = 59^\circ
\]
\[
m\angle FDE = 5x - 2 = 5(15) - 2 = 75 - 2 = 73^\circ
\]
\[
m\angle CDE = 10x - 18 = 10(15) - 18 = 150 - 18 = 132^\circ
\]

Thus, the measures are:
\[
\boxed{59^\circ, 73^\circ, 132^\circ}
\]

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Problem 7

Given:
- \( m\angle LMP \) is 11 degrees more than \( m\angle NMP \)
- \( m\angle NML = 137^\circ \)

Let \( m\angle NMP = y \). Then:
\[
m\angle LMP = y + 11
\]

Since \( \angle NML \) is the sum of \( \angle NMP \) and \( \angle LMP \):
\[
m\angle NML = m\angle NMP + m\angle LMP
\]
Substitute the given values:
\[
137 = y + (y + 11)
\]
Simplify:
\[
137 = 2y + 11
\]
Isolate \( y \):
\[
137 - 11 = 2y
\]
\[
126 = 2y
\]
\[
y = 63
\]

Now, find \( m\angle LMP \):
\[
m\angle LMP = y + 11 = 63 + 11 = 74^\circ
\]

Thus, the measures are:
\[
\boxed{74^\circ, 63^\circ}
\]

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Problem 8

Given:
- \( m\angle ABC \) is one degree less than three times \( m\angle ABD \)
- \( m\angle DBC = 47^\circ \)

Let \( m\angle ABD = z \). Then:
\[
m\angle ABC = 3z - 1
\]

Since \( \angle ABC \) is the sum of \( \angle ABD \) and \( \angle DBC \):
\[
m\angle ABC = m\angle ABD + m\angle DBC
\]
Substitute the given values:
\[
3z - 1 = z + 47
\]
Isolate \( z \):
\[
3z - z = 47 + 1
\]
\[
2z = 48
\]
\[
z = 24
\]

Now, find \( m\angle ABC \):
\[
m\angle ABC = 3z - 1 = 3(24) - 1 = 72 - 1 = 71^\circ
\]

Thus, the measures are:
\[
\boxed{24^\circ, 71^\circ}
\]

---

Problem 9

Given:
- \( OQ \) bisects \( \angle POR \)
- \( m\angle QOR = (2x - 7)^\circ \)
- \( m\angle POQ = (x + 3)^\circ \)
- \( m\angle POR = 112^\circ \)

Since \( OQ \) bisects \( \angle POR \), we have:
\[
m\angle POQ = m\angle QOR
\]
Substitute the given expressions:
\[
x + 3 = 2x - 7
\]
Isolate \( x \):
\[
x - 2x = -7 - 3
\]
\[
-x = -10
\]
\[
x = 10
\]

Now, substitute \( x = 10 \) back into the expressions for each angle:
\[
m\angle POQ = x + 3 = 10 + 3 = 13^\circ
\]
\[
m\angle QOR = 2x - 7 = 2(10) - 7 = 20 - 7 = 13^\circ
\]
\[
m\angle POR = 112^\circ \quad \text{(given)}
\]

Thus, the measures are:
\[
\boxed{13^\circ, 13^\circ, 112^\circ}
\]

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Problem 10

Given:
- \( CD \) is a diameter of a circle
- \( m\angle GED = (9x - 17)^\circ \)
- \( m\angle EDB = (4x + 12)^\circ \)
- \( m\angle CFD = 47^\circ \)

Since \( CD \) is a diameter, \( \angle GED \) and \( \angle EDB \) are inscribed angles that intercept the same arc \( \widehat{CE} \). Therefore:
\[
m\angle GED = m\angle EDB
\]
Substitute the given expressions:
\[
9x - 17 = 4x + 12
\]
Isolate \( x \):
\[
9x - 4x = 12 + 17
\]
\[
5x = 29
\]
\[
x = \frac{29}{5}
\]

Now, substitute \( x = \frac{29}{5} \) back into the expressions for each angle:
\[
m\angle GED = 9x - 17 = 9\left(\frac{29}{5}\right) - 17 = \frac{261}{5} - 17 = \frac{261}{5} - \frac{85}{5} = \frac{176}{5} = 35.2^\circ
\]
\[
m\angle EDB = 4x + 12 = 4\left(\frac{29}{5}\right) + 12 = \frac{116}{5} + 12 = \frac{116}{5} + \frac{60}{5} = \frac{176}{5} = 35.2^\circ
\]
\[
m\angle CFD = 47^\circ \quad \text{(given)}
\]

Thus, the measures are:
\[
\boxed{35.2^\circ, 35.2^\circ, 47^\circ}
\]

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Final Answers:

1. Problem 6: \(\boxed{59^\circ, 73^\circ, 132^\circ}\)
2. Problem 7: \(\boxed{74^\circ, 63^\circ}\)
3. Problem 8: \(\boxed{24^\circ, 71^\circ}\)
4. Problem 9: \(\boxed{13^\circ, 13^\circ, 112^\circ}\)
5. Problem 10: \(\boxed{35.2^\circ, 35.2^\circ, 47^\circ}\)