Question

geometry test review (unit 1.4 - 1.5) 1) $mangle nhg = 3x + 10$, $mangle ihn=90^{circ}$, and $mangle ihg = 14x + 1$. find $x$. 2) $mangle srq = 71^{circ}$, $mangle grq = 4x - 10$, and $mangle srg = 3x - 3$. find $x$. 3) $mangle trq=x + 31$, $mangle srt=x + 135$, and $mangle srq = 156^{circ}$. find $x$. 4) $mangle tul = 5 + 10x$, $mangle luv = 16x + 2$, and $mangle tuv = 163^{circ}$. find $x$. 5) find $x$ if $mangle efn = 1+7x$, $mangle nfg = 28x$, and $mangle efg = 176^{circ}$. 6) $mangle fkl = x + 156$, $mangle jkf = x + 3$, and $mangle jkl = 167^{circ}$. find $mangle fkl$. 7) $mangle abi = 2x + 12$, $mangle ibc = 150^{circ}$, and $mangle abc = 24x + 8$. find $mangle abc$. 8) $mangle pqr = 55x + 2$, $mangle pqf$, and $mangle fqr = 65^{circ}$. find $m$

Explanation:

Step1: Substitute angle - measures into the equation

\(14x + 1=90+(3x + 10)\)
## Step2: Simplify the right - hand side of the equation
\(14x + 1=3x+100\)
## Step3: Subtract \(3x\) from both sides
\(14x-3x + 1=3x-3x + 100\), which gives \(11x+1 = 100\)
## Step4: Subtract 1 from both sides
\(11x+1 - 1=100 - 1\), so \(11x=99\)
## Step5: Divide both sides by 11
\(x=\frac{99}{11}=9\)
- # Answer:
\(x = 9\)
2. **For the second problem**:
- Given \(m\angle SRQ = 71^{\circ}\), \(m\angle GRQ = 4x-10\), and \(m\angle SRG = 3x - 3\).
- Since \(\angle SRQ=\angle SRG+\angle GRQ\), we set up the equation:
- # Explanation:
## Step1: Substitute angle - measures into the equation
\(71=(3x - 3)+(4x-10)\)
## Step2: Combine like terms on the right - hand side
\(71 = 7x-13\)
## Step3: Add 13 to both sides
\(71 + 13=7x-13 + 13\), so \(84 = 7x\)
## Step4: Divide both sides by 7
\(x=\frac{84}{7}=12\)
- # Answer:
\(x = 12\)
3. **For the third problem**:
- Given \(m\angle TRQ=x + 31\), \(m\angle SRT=x + 135\), and \(m\angle SRQ = 156^{\circ}\).
- Since \(\angle SRQ=\angle SRT+\angle TRQ\), we set up the equation:
- # Explanation:
## Step1: Substitute angle - measures into the equation
\(156=(x + 135)+(x + 31)\)
## Step2: Combine like terms on the right - hand side
\(156=2x+166\)
## Step3: Subtract 166 from both sides
\(156-166=2x+166 - 166\), so \(- 10=2x\)
## Step4: Divide both sides by 2
\(x=-5\)
- # Answer:
\(x=-5\)
4. **For the fourth problem**:
- Given \(m\angle TUL = 5 + 10x\), \(m\angle LUV = 16x+2\), and \(m\angle TUV = 163^{\circ}\).
- Since \(\angle TUV=\angle TUL+\angle LUV\), we set up the equation:
- # Explanation:
## Step1: Substitute angle - measures into the equation
\(163=(5 + 10x)+(16x+2)\)
## Step2: Combine like terms on the right - hand side
\(163=26x + 7\)
## Step3: Subtract 7 from both sides
\(163-7=26x+7 - 7\), so \(156=26x\)
## Step4: Divide both sides by 26
\(x = 6\)
- # Answer:
\(x = 6\)
5. **For the fifth problem**:
- Given \(m\angle EFN = 1+7x\), \(m\angle NFG = 28x\), and \(m\angle EFG = 176^{\circ}\).
- Since \(\angle EFG=\angle EFN+\angle NFG\), we set up the equation:
- # Explanation:
## Step1: Substitute angle - measures into the equation
\(176=(1 + 7x)+28x\)
## Step2: Combine like terms on the right - hand side
\(176=35x + 1\)
## Step3: Subtract 1 from both sides
\(176-1=35x+1 - 1\), so \(175=35x\)
## Step4: Divide both sides by 35
\(x = 5\)
- # Answer:
\(x = 5\)
6. **For the sixth problem**:
- Given \(m\angle FKL=x + 156\), \(m\angle JKF=x + 3\), and \(m\angle JKL = 167^{\circ}\).
- Since \(\angle JKL=\angle JKF+\angle FKL\), we set up the equation:
- # Explanation:
## Step1: Substitute angle - measures into the equation
\(167=(x + 3)+(x + 156)\)
## Step2: Combine like terms on the right - hand side
\(167=2x+159\)
## Step3: Subtract 159 from both sides
\(167-159=2x+159 - 159\), so \(8 = 2x\)
## Step4: Divide both sides by 2
\(x = 4\)
- Then find \(m\angle FKL\):
\(m\angle FKL=x + 156\), substituting \(x = 4\), we get \(m\angle FKL=4 + 156=160^{\circ}\)
- # Answer:
\(m\angle FKL = 160^{\circ}\)
7. **For the seventh problem**:
- Given \(m\angle ABI = 2x+12\), \(m\angle IBC = 150^{\circ}\), and \(m\angle ABC = 24x + 8\).
- Since \(\angle ABC=\angle ABI+\angle IBC\), we set up the equation:
- # Explanation:
## Step1: Substitute angle - measures into the equation
\(24x + 8=(2x+12)+150\)
## Step2: Combine like terms on the right - hand side
\(24x + 8=2x+162\)
## Step3: Subtract \(2x\) from both sides
\(24x-2x + 8=2x-2x + 162\), so \(22x+8 = 162\)
## Step4: Subtract 8 from both sides
\(22x+8 - 8=162 - 8\), so \(22x=154\)
## Step5: Divide both sides by 22
\(x = 7\)
- Then find \(m\angle ABC\):
\(m\angle ABC=24x + 8\), substituting \(x = 7\), we get \(m\angle ABC=24\times7 + 8=168 + 8=176^{\circ}\)
- # Answer:
\(m\angle ABC = 176^{\circ}\)
8. **For the eighth problem**:
- Since the problem is incomplete (no full information about the relationship between \(\angle PQR\), \(\angle PQF\), and \(\angle FQR\) is given), we cannot solve it.

So, the solutions for the first 7 solvable problems are:
1. \(x = 9\)
2. \(x = 12\)
3. \(x=-5\)
4. \(x = 6\)
5. \(x = 5\)
6. \(m\angle FKL = 160^{\circ}\)
7. \(m\angle ABC = 176^{\circ}\)

Answer:

1. For the first problem:

• Given \(m\angle NHG = 3x + 10\), \(m\angle IHN=90^{\circ}\), and \(m\angle IHG = 14x + 1\).
• Since \(\angle IHG=\angle IHN+\angle NHG\), we can set up the equation:
• # Explanation:

Step1: Substitute angle - measures into the equation

\(14x + 1=90+(3x + 10)\)

Step2: Simplify the right - hand side of the equation

\(14x + 1=3x+100\)

Step3: Subtract \(3x\) from both sides

\(14x-3x + 1=3x-3x + 100\), which gives \(11x+1 = 100\)

Step4: Subtract 1 from both sides

\(11x+1 - 1=100 - 1\), so \(11x=99\)

Step5: Divide both sides by 11

\(x=\frac{99}{11}=9\)

• # Answer:

\(x = 9\)

1. For the second problem:

• Given \(m\angle SRQ = 71^{\circ}\), \(m\angle GRQ = 4x-10\), and \(m\angle SRG = 3x - 3\).
• Since \(\angle SRQ=\angle SRG+\angle GRQ\), we set up the equation:
• # Explanation:

Step1: Substitute angle - measures into the equation

\(71=(3x - 3)+(4x-10)\)

Step2: Combine like terms on the right - hand side

\(71 = 7x-13\)

Step3: Add 13 to both sides

\(71 + 13=7x-13 + 13\), so \(84 = 7x\)

Step4: Divide both sides by 7

\(x=\frac{84}{7}=12\)

• # Answer:

\(x = 12\)

1. For the third problem:

• Given \(m\angle TRQ=x + 31\), \(m\angle SRT=x + 135\), and \(m\angle SRQ = 156^{\circ}\).
• Since \(\angle SRQ=\angle SRT+\angle TRQ\), we set up the equation:
• # Explanation:

Step1: Substitute angle - measures into the equation

\(156=(x + 135)+(x + 31)\)

Step2: Combine like terms on the right - hand side

\(156=2x+166\)

Step3: Subtract 166 from both sides

\(156-166=2x+166 - 166\), so \(- 10=2x\)

Step4: Divide both sides by 2

\(x=-5\)

• # Answer:

\(x=-5\)

1. For the fourth problem:

• Given \(m\angle TUL = 5 + 10x\), \(m\angle LUV = 16x+2\), and \(m\angle TUV = 163^{\circ}\).
• Since \(\angle TUV=\angle TUL+\angle LUV\), we set up the equation:
• # Explanation:

Step1: Substitute angle - measures into the equation

\(163=(5 + 10x)+(16x+2)\)

Step2: Combine like terms on the right - hand side

\(163=26x + 7\)

Step3: Subtract 7 from both sides

\(163-7=26x+7 - 7\), so \(156=26x\)

Step4: Divide both sides by 26

\(x = 6\)

• # Answer:

\(x = 6\)

1. For the fifth problem:

• Given \(m\angle EFN = 1+7x\), \(m\angle NFG = 28x\), and \(m\angle EFG = 176^{\circ}\).
• Since \(\angle EFG=\angle EFN+\angle NFG\), we set up the equation:
• # Explanation:

Step1: Substitute angle - measures into the equation

\(176=(1 + 7x)+28x\)

Step2: Combine like terms on the right - hand side

\(176=35x + 1\)

Step3: Subtract 1 from both sides

\(176-1=35x+1 - 1\), so \(175=35x\)

Step4: Divide both sides by 35

\(x = 5\)

• # Answer:

\(x = 5\)

1. For the sixth problem:

• Given \(m\angle FKL=x + 156\), \(m\angle JKF=x + 3\), and \(m\angle JKL = 167^{\circ}\).
• Since \(\angle JKL=\angle JKF+\angle FKL\), we set up the equation:
• # Explanation:

Step1: Substitute angle - measures into the equation

\(167=(x + 3)+(x + 156)\)

Step2: Combine like terms on the right - hand side

\(167=2x+159\)

Step3: Subtract 159 from both sides

\(167-159=2x+159 - 159\), so \(8 = 2x\)

Step4: Divide both sides by 2

\(x = 4\)

• Then find \(m\angle FKL\):

\(m\angle FKL=x + 156\), substituting \(x = 4\), we get \(m\angle FKL=4 + 156=160^{\circ}\)

• # Answer:

\(m\angle FKL = 160^{\circ}\)

1. For the seventh problem:

• Given \(m\angle ABI = 2x+12\), \(m\angle IBC = 150^{\circ}\), and \(m\angle ABC = 24x + 8\).
• Since \(\angle ABC=\angle ABI+\angle IBC\), we set up the equation:
• # Explanation:

Step1: Substitute angle - measures into the equation

\(24x + 8=(2x+12)+150\)

Step2: Combine like terms on the right - hand side

\(24x + 8=2x+162\)

Step3: Subtract \(2x\) from both sides

\(24x-2x + 8=2x-2x + 162\), so \(22x+8 = 162\)

Step4: Subtract 8 from both sides

\(22x+8 - 8=162 - 8\), so \(22x=154\)

Step5: Divide both sides by 22

\(x = 7\)

• Then find \(m\angle ABC\):

\(m\angle ABC=24x + 8\), substituting \(x = 7\), we get \(m\angle ABC=24\times7 + 8=168 + 8=176^{\circ}\)

• # Answer:

\(m\angle ABC = 176^{\circ}\)

1. For the eighth problem:

• Since the problem is incomplete (no full information about the relationship between \(\angle PQR\), \(\angle PQF\), and \(\angle FQR\) is given), we cannot solve it.

So, the solutions for the first 7 solvable problems are:

1. \(x = 9\)
2. \(x = 12\)
3. \(x=-5\)
4. \(x = 6\)
5. \(x = 5\)
6. \(m\angle FKL = 160^{\circ}\)
7. \(m\angle ABC = 176^{\circ}\)