\(m\angle NKL = 8(\quad) - 30 =\) _______
Part d: What is the \(m\angle JKN\)? (Hint: \(\angle JKN\) and \(\angle NKL\) form a straight line = \(180^\circ\))
Equation: \(m\angle JKN + m\angle NKL = 180^\circ\) \(m\angle JKN =\) ________\(^\circ\)
4
\(m\angle CXD = 90^\circ\) and ray \(\vec{XE}\) intersects inside \(\angle CXD\). If \(m\angle CXE = (2x)^\circ\) and \(m\angle EXD = (3x + 15)^\circ\), what is \(m\angle CXE\)?
Step 1: Draw & Label the Diagram
Draw a right angle, add ray XE inside
Step 2: Set Up \(2x + (3x + 15) = 90^\circ\) & Solve
\(x =\) _____ \(m\angle CXE = 2(x) =\) ______\(^\circ\)
Geometry • Angle Bisectors Practice Page 2 of 3
10th Grade Geometry • Challenge & Mastery
Name: Date:
LEVEL 3 Challenge & Exit Ticket Mastery
Opposite rays + Quadratic angle equations
5
In the figure, \(\vec{XA}\) and \(\vec{XE}\) are opposite rays (forming straight angle \(= 180^\circ\)), and \(\angle AXC\) is bisected by \(\vec{XB}\). If \(m\angle BXC = (8x - 30)^\circ\) and \(m\angle AXB = (4x + 10)^\circ\), find \(m\angle BXE\).
X A E B C
Step 1: Set Halves Equal \(8x - 30 = 4x + 10\)
\(x =\) _____
Step 2: Find \(m\angle AXB\) Plug \(x\) into \(4x + 10\)
\(m\angle AXB =\) _____\(^\circ\)
Step 3: Find \(m\angle BXE\) Straight angle \(= 180^\circ - m\angle AXB\)
\(m\angle BXE =\) _____\(^\circ\)
6
Ray \(\vec{NL}\) bisects \(\angle MNK\). The total whole angle is \(m\angle KNM = (x^2 - 9)^\circ\) and one half is \(m\angle LNK = (4x)^\circ\). Determine the measure of \(m\angle LNM\).
Template \(2(\text{Half}) = \text{Whole}\)
Work Space: Set up \(2(4x) = x^2 - 9\) → Rearrange to \(x^2 - 8x - 9 = 0\)
Factored: \((x - 9)(x + 1) = 0\) → Positive solution: \(x = 9\) (angle cannot be negative)
Determine \(m\angle LNM\):
[ ] A. \(9^\circ\)
[ ] B. \(36^\circ\)
[ ] C. \(72^\circ\)
[ ] D. \(4^\circ\)
*Half = \(4(9) = 36^\circ\)
Self-Assessment: How confident do you feel using angle bisectors?
👍 Got It! ✋ Almost There 👎 Need Help
One question or reminder for myself:
Geometry • Angle Bisectors Practice Page 3 of 3
Teacher Solutions Guide • Angle Bisector Mastery Page 1 of 2
Level 3 • Challenge Solutions & SPED Strategies
Mastery Key
LEVEL 3
\(m\angle BXE = 130^\circ\)
Step 1: Bisector Equate \(8x - 30 = 4x + 10\)
\(4x = 40 \implies \mathbf{x = 10}\)
Step 2: Find Angle AXB \(m\angle AXB = 4(10) + 10\)
\(= 40 + 10 = \mathbf{50^\circ}\)
Step 3: Linear Pair Rule \(m\angle BXE = 180^\circ - 50^\circ\)
\(= \mathbf{130^\circ}\)
*Alternative method: \(m\angle AXC = 50 + 50 = 100^\circ\). Then \(m\angle CXE = 180 - 100 = 80^\circ\). Then \(m\angle BXE = 50^\circ + 80^\circ = 130^\circ\). Both confirm 130°!
EXIT TICKET
Option B: \(36^\circ\)
1. Template Setup: One half given (\(4x\)) and whole angle given (\(x^2 - 9\)) → \(2(\text{Half}) = \text{Whole}\)
\(2(4x) = x^2 - 9 \implies 8x = x^2 - 9 \implies x^2 - 8x - 9 = 0\)
2. Factor: \((x - 9)(x + 1) = 0 \implies x = 9\) or \(x = -1\).
3. Reject extraneous negative: If \(x = -1\), \(4(-1) = -4^\circ\) (angles cannot be negative). Thus, \(x = 9\).
4. Calculate requested angle: The problem asks for \(m\angle LNM\) (the other half!). Since \(\vec{NL}\) bisects, \(m\angle LNM = m\angle LNK = 4(9) = \mathbf{36^\circ}\). (Whole angle is \(9^2 - 9 = 72^\circ\)).
PDF Problem #3 Solution: \(m\angle JKN = 5x^2 - 3\), \(m\angle LKN = -14x\)
\(5x^2 - 3 = -14x \implies 5x^2 + 14x - 3 = 0\)
\((5x - 1)(x + 3) = 0 \implies x = -3\)
\(m\angle JKN = -14(-3) = \mathbf{42^\circ}\)
Exit Ticket #2 Solution: Ray \(\vec{KM}\) bisects \(\angle JKN\). Which is true?
Correct: Option (b) \(m\angle JKM = m\angle MKN\)
By definition, bisector creates two congruent angles.
Color-Coding Have students highlight the bisector ray in yellow and draw identical colored arc tick marks on the two halves.
Physical Covering Use a sticky note to cover the non-bisected portion of the straight line to prevent visual overload during equation setup.
Target Question Box Have students highlight what the question asks for (\(x\) vs angle) before writing anything.
Teacher Solutions Guide • Angle Bisector Mastery Page 2 of 2
