Question

1 (g.5b) in the diagram below, assume that line m is a straight - line. if m∠3 = 47°, what is the m∠1? record your answer and fill in the bubbles on your answer document. be sure to use the correct place value.
2 (g.2a, g.2b) n is the mid - point of jl. if jn=x and nl = 3x + 3. find jl. record your answer and fill in the bubbles on your answer document. be sure to use the correct place value.
3 (g.5b) bd bisects ∠abc. if m∠abd=(7x - 1)° and m∠dbc=(4x + 8)°, find m∠abd. record your answer and fill in the bubbles on your answer document. be sure to use the correct place value.
4 (g.5b) in the figure below, what is the value of y? record your answer and fill in the bubbles on your answer document. be sure to use the correct place value.

Explanation:

1. Problem 1:

• # Explanation:
• ## Step1: Use linear - pair property
• Since \(m\angle1\) and \(m\angle2\) form a linear pair and \(m\angle2 = 47^{\circ}\), and the sum of angles in a linear pair is \(180^{\circ}\). Let \(m\angle1=x\), then \(x + 47^{\circ}=180^{\circ}\).
• ## Step2: Solve for \(m\angle1\)
• \(x=180^{\circ}-47^{\circ}=133^{\circ}\).
• # Answer:
• \(133\)

1. Problem 2:

• # Explanation:
• ## Step1: Use mid - point property
• If \(N\) is the mid - point of \(\overline{JL}\), then \(JN = NL\). Given \(JN = 6x\) and \(NL=3x + 3\), we set up the equation \(6x=3x + 3\).
• ## Step2: Solve for \(x\)
• Subtract \(3x\) from both sides: \(6x-3x=3x + 3-3x\), so \(3x=3\), and \(x = 1\).
• ## Step3: Find \(JL\)
• Since \(JL=JN + NL\) and \(JN = 6x\), \(NL=3x + 3\), substituting \(x = 1\), we get \(JN=6\times1 = 6\), \(NL=3\times1+3=6\). Then \(JL=6 + 6=12\).
• # Answer:
• \(12\)

1. Problem 3:

• # Explanation:
• ## Step1: Use angle - bisector property
• Since \(\overrightarrow{BD}\) bisects \(\angle ABC\), then \(m\angle ABD=m\angle DBC\). Set up the equation \(7x-1=4x + 8\).
• ## Step2: Solve for \(x\)
• Subtract \(4x\) from both sides: \(7x-4x-1=4x-4x + 8\), so \(3x-1=8\). Add \(1\) to both sides: \(3x-1 + 1=8 + 1\), then \(3x=9\), and \(x = 3\).
• ## Step3: Find \(m\angle ABD\)
• Substitute \(x = 3\) into \(m\angle ABD=7x-1\). So \(m\angle ABD=7\times3-1=21 - 1=20^{\circ}\).
• # Answer:
• \(20\)

1. Problem 4:

• # Explanation:
• ## Step1: Use vertical - angles and linear - pair property
• The angles \(4x^{\circ}\) and \(2x^{\circ}\) are adjacent angles that form a straight - line, so \(4x+2x = 180\) (since the sum of angles on a straight - line is \(180^{\circ}\)).
• ## Step2: Solve for \(x\)
• Combine like terms: \(6x=180\), then \(x = 30\).
• ## Step3: Find \(y\)
• The angle \(y^{\circ}\) and \(4x^{\circ}\) are vertical angles, so \(y = 4x\). Substitute \(x = 30\) into \(y = 4x\), we get \(y=4\times30=120\).
• # Answer:
• \(120\)

Answer:

1. Problem 1:

• # Explanation:
• ## Step1: Use linear - pair property
• Since \(m\angle1\) and \(m\angle2\) form a linear pair and \(m\angle2 = 47^{\circ}\), and the sum of angles in a linear pair is \(180^{\circ}\). Let \(m\angle1=x\), then \(x + 47^{\circ}=180^{\circ}\).
• ## Step2: Solve for \(m\angle1\)
• \(x=180^{\circ}-47^{\circ}=133^{\circ}\).
• # Answer:
• \(133\)

1. Problem 2:

• # Explanation:
• ## Step1: Use mid - point property
• If \(N\) is the mid - point of \(\overline{JL}\), then \(JN = NL\). Given \(JN = 6x\) and \(NL=3x + 3\), we set up the equation \(6x=3x + 3\).
• ## Step2: Solve for \(x\)
• Subtract \(3x\) from both sides: \(6x-3x=3x + 3-3x\), so \(3x=3\), and \(x = 1\).
• ## Step3: Find \(JL\)
• Since \(JL=JN + NL\) and \(JN = 6x\), \(NL=3x + 3\), substituting \(x = 1\), we get \(JN=6\times1 = 6\), \(NL=3\times1+3=6\). Then \(JL=6 + 6=12\).
• # Answer:
• \(12\)

1. Problem 3:

• # Explanation:
• ## Step1: Use angle - bisector property
• Since \(\overrightarrow{BD}\) bisects \(\angle ABC\), then \(m\angle ABD=m\angle DBC\). Set up the equation \(7x-1=4x + 8\).
• ## Step2: Solve for \(x\)
• Subtract \(4x\) from both sides: \(7x-4x-1=4x-4x + 8\), so \(3x-1=8\). Add \(1\) to both sides: \(3x-1 + 1=8 + 1\), then \(3x=9\), and \(x = 3\).
• ## Step3: Find \(m\angle ABD\)
• Substitute \(x = 3\) into \(m\angle ABD=7x-1\). So \(m\angle ABD=7\times3-1=21 - 1=20^{\circ}\).
• # Answer:
• \(20\)

1. Problem 4:

• # Explanation:
• ## Step1: Use vertical - angles and linear - pair property
• The angles \(4x^{\circ}\) and \(2x^{\circ}\) are adjacent angles that form a straight - line, so \(4x+2x = 180\) (since the sum of angles on a straight - line is \(180^{\circ}\)).
• ## Step2: Solve for \(x\)
• Combine like terms: \(6x=180\), then \(x = 30\).
• ## Step3: Find \(y\)
• The angle \(y^{\circ}\) and \(4x^{\circ}\) are vertical angles, so \(y = 4x\). Substitute \(x = 30\) into \(y = 4x\), we get \(y=4\times30=120\).
• # Answer:
• \(120\)