Question

1. given: < 1 = 7x - 1 < 2 = 11x + 1 < 1 and < 2 are complementary find: the measure of each angle. 2. given: < 1 = 4x + 3 < 2 = x + 17 < 1 and < 2 form a linear pair find: the measure of each angle

Explanation:

Step1: Use the property of complementary angles

Complementary angles sum to \(90^{\circ}\). So, \(\angle1+\angle2 = 90\). Substitute \(\angle1 = 7x - 1\) and \(\angle2=11x + 1\) into the equation:
\((7x - 1)+(11x + 1)=90\)
Simplify the left - hand side:
\(7x-1 + 11x+1=90\)
\(18x=90\)
Solve for \(x\):
\(x=\frac{90}{18}=5\)

Step2: Find the measure of \(\angle1\)

Substitute \(x = 5\) into \(\angle1=7x - 1\):
\(\angle1=7\times5-1=35 - 1=34^{\circ}\)

Step3: Find the measure of \(\angle2\)

Substitute \(x = 5\) into \(\angle2=11x + 1\):
\(\angle2=11\times5+1=55 + 1=56^{\circ}\)

Step1: Use the property of linear - pair angles

Linear - pair angles sum to \(180^{\circ}\). So, \(\angle1+\angle2 = 180\). Substitute \(\angle1 = 4x + 3\) and \(\angle2=x + 17\) into the equation:
\((4x + 3)+(x + 17)=180\)
Simplify the left - hand side:
\(4x+3+x + 17=180\)
\(5x+20=180\)
Subtract 20 from both sides:
\(5x=180 - 20=160\)
Solve for \(x\):
\(x=\frac{160}{5}=32\)

Step2: Find the measure of \(\angle1\)

Substitute \(x = 32\) into \(\angle1=4x + 3\):
\(\angle1=4\times32+3=128 + 3=131^{\circ}\)

Step3: Find the measure of \(\angle2\)

Substitute \(x = 32\) into \(\angle2=x + 17\):
\(\angle2=32+17=49^{\circ}\)

Answer:

\(\angle1 = 34^{\circ}\), \(\angle2 = 56^{\circ}\)