Question

1. ∠1 and ∠2 form a linear pair. if m∠1 = (5x + 9)° and m∠2 = (3x + 11)°, find the measure of each angle.
2. ∠1 and ∠2 are vertical angles. if m∠1 = (17x + 1)° and m∠2 = (20x - 14)°, find m∠2

Explanation:

Step1: Use the property of linear pair (for problem 6)

Since \(\angle1\) and \(\angle2\) form a linear pair, \(m\angle1 + m\angle2=180^{\circ}\).
So, \((5x + 9)+(3x + 11)=180\).
Combine like - terms: \(5x+3x+9 + 11=180\), \(8x+20 = 180\).
Subtract 20 from both sides: \(8x=180 - 20\), \(8x=160\).
Divide both sides by 8: \(x = 20\).
Then \(m\angle1=(5x + 9)^{\circ}=(5\times20+9)^{\circ}=109^{\circ}\) and \(m\angle2=(3x + 11)^{\circ}=(3\times20 + 11)^{\circ}=71^{\circ}\).

Step2: Use the property of vertical angles (for problem 7)

Since \(\angle1\) and \(\angle2\) are vertical angles, \(m\angle1=m\angle2\).
So, \(17x + 1=20x-14\).
Subtract \(17x\) from both sides: \(1=20x-17x-14\).
Simplify: \(1 = 3x-14\).
Add 14 to both sides: \(1+14=3x\), \(15 = 3x\).
Divide both sides by 3: \(x = 5\).
Then \(m\angle2=(20x-14)^{\circ}=(20\times5-14)^{\circ}=86^{\circ}\).

Answer:

For problem 6: \(m\angle1 = 109^{\circ}\), \(m\angle2=71^{\circ}\).
For problem 7: \(m\angle2 = 86^{\circ}\).