Question

geometry test a (unit 1.4 - 1.5) 1) (mangle gqr = 17x + 4), (mangle pqg=3x + 6), and (mangle pqr = 170^{circ}). find (x). 2) (mangle duv=x + 116), (mangle tud=x + 39), and (mangle tuv = 143^{circ}). find (x). 3) (mangle lmj=x + 28), (mangle jmn=x + 124), and (mangle lmn = 144^{circ}). find (mangle jmn). 4) (mangle kls = 5x-7), (mangle slm = 133^{circ}), and (mangle klm = 18x + 9). find (mangle) classify each angle as acute, obtuse, right, or straight.

Explanation:

1. For the first problem:

• Given \(m\angle GQR = 17x + 4\), \(m\angle PQG=3x + 6\), and \(m\angle PQR = 170^{\circ}\).
• Since \(\angle PQR=\angle PQG+\angle GQR\) (angle - addition postulate), we can set up the equation:
• Step1: Write the equation
• \((3x + 6)+(17x + 4)=170\).
• Step2: Combine like - terms
• \(3x+17x+6 + 4=170\), which simplifies to \(20x+10 = 170\).
• Step3: Isolate the variable term
• Subtract 10 from both sides of the equation: \(20x=170 - 10\), so \(20x=160\).
• Step4: Solve for \(x\)
• Divide both sides by 20: \(x=\frac{160}{20}=8\).

1. For the second problem:

• Given \(m\angle DUV=x + 116\), \(m\angle TUD=x + 39\), and \(m\angle TUV = 143^{\circ}\).
• Since \(\angle TUV=\angle TUD+\angle DUV\) (angle - addition postulate), we set up the equation:
• Step1: Write the equation
• \((x + 39)+(x + 116)=143\).
• Step2: Combine like - terms
• \(x+x+39 + 116=143\), which simplifies to \(2x+155 = 143\).
• Step3: Isolate the variable term
• Subtract 155 from both sides: \(2x=143 - 155=-12\).
• Step4: Solve for \(x\)
• Divide both sides by 2: \(x=\frac{-12}{2}=-6\).

1. For the third problem:

• Given \(m\angle LMJ=x + 28\), \(m\angle JMN=x + 124\), and \(m\angle LMN = 144^{\circ}\).
• Since \(\angle LMN=\angle LMJ+\angle JMN\) (angle - addition postulate), we set up the equation:
• Step1: Write the equation
• \((x + 28)+(x + 124)=144\).
• Step2: Combine like - terms
• \(x+x+28 + 124=144\), which simplifies to \(2x+152 = 144\).
• Step3: Isolate the variable term
• Subtract 152 from both sides: \(2x=144 - 152=-8\).
• Step4: Solve for \(x\)
• Divide both sides by 2: \(x=\frac{-8}{2}=-4\).
• Step5: Find \(m\angle JMN\)
• Substitute \(x=-4\) into the expression for \(m\angle JMN\): \(m\angle JMN=x + 124=-4+124 = 120^{\circ}\).

1. For the fourth problem:

• Given \(m\angle KLS=5x - 7\), \(m\angle SLM = 133^{\circ}\), and \(m\angle KLM=18x + 9\).
• Since \(\angle KLM=\angle KLS+\angle SLM\) (angle - addition postulate), we set up the equation:
• Step1: Write the equation
• \((5x - 7)+133=18x + 9\).
• Step2: Combine like - terms on the left - hand side
• \(5x+126 = 18x + 9\).
• Step3: Move the \(x\) terms to one side
• Subtract \(5x\) from both sides: \(126=18x - 5x+9\), which simplifies to \(126 = 13x+9\).
• Step4: Isolate the variable term
• Subtract 9 from both sides: \(13x=126 - 9 = 117\).
• Step5: Solve for \(x\)
• Divide both sides by 13: \(x=\frac{117}{13}=9\).
• Step6: Find \(m\angle KLS\)
• Substitute \(x = 9\) into the expression for \(m\angle KLS\): \(m\angle KLS=5x - 7=5\times9 - 7=45 - 7 = 38^{\circ}\).

Answer:

1. \(x = 8\)
2. \(x=-6\)
3. \(m\angle JMN = 120^{\circ}\)
4. \(x = 9\)