Question

#13 m∠mnq =
(2x - 1)°
(5x - 40)°

Explanation:

1. **For the first - angle problem (with angles at point \(K\)):

• Assume that \(\angle GKJ\) and \(\angle HKG\) are complementary or supplementary. Since no information about the relationship between the angles is given, if we assume they are adjacent and form a straight - line (supplementary, i.e., \(\angle GKJ+\angle HKG = 180^{\circ}\)), we can set up the equation:
• # Explanation:

Step1: Set up the equation based on the angle - sum property

\((13x - 1)+(9x + 3)=180\)

Step2: Combine like terms

\(13x+9x-1 + 3=180\), which simplifies to \(22x+2 = 180\)

Step3: Isolate the variable term

Subtract 2 from both sides: \(22x=180 - 2=178\)

Step4: Solve for \(x\)

Divide both sides by 22: \(x=\frac{178}{22}=\frac{89}{11}\approx8.09\)

1. **For the second - angle problem (with angles at point \(N\)):

• Assume that \(\angle MNQ\) and \(\angle PNR\) are vertical angles (vertical angles are equal).
• # Explanation:

Step1: Set up the equation based on the vertical - angle property

\(2x-1 = 5x - 40\)

Step2: Move the \(x\) terms to one side

Subtract \(2x\) from both sides: \(-1=5x-2x - 40\), which simplifies to \(-1 = 3x-40\)

Step3: Isolate the variable term

Add 40 to both sides: \(3x=-1 + 40=39\)

Step4: Solve for \(x\)

Divide both sides by 3: \(x = 13\)
Then, to find \(m\angle MNQ\), substitute \(x = 13\) into the expression for \(\angle MNQ\):
\(m\angle MNQ=2x-1=2\times13-1=26 - 1=25^{\circ}\)

Answer:

For the first problem, \(x=\frac{89}{11}\) (assuming supplementary angles). For the second problem, \(m\angle MNQ = 25^{\circ}\)