Question

∠a and ∠b are supplementary angles. find m∠a and m∠b.

1. m∠a=(x + 50)°, m∠b=(x + 100)°
2. m∠a = 6x°, m∠b=(x + 5)°
3. m∠a=(2x + 3)°, m∠b=(3x - 223)°
4. m∠a=(-4x + 40)°, m∠b=(x + 50)°

Explanation:

Step1: Recall supplementary - angle property

Supplementary angles add up to 180°. So, for each pair of angles ∠A and ∠B, we set up the equation \(m\angle A + m\angle B=180^{\circ}\).

Step2: Solve for \(x\) in problem 29

Given \(m\angle A=(x + 50)^{\circ}\) and \(m\angle B=(x + 100)^{\circ}\), we substitute into the equation:
\((x + 50)+(x + 100)=180\).
Combine like - terms: \(2x+150 = 180\).
Subtract 150 from both sides: \(2x=180 - 150=30\).
Divide both sides by 2: \(x = 15\).
Then \(m\angle A=x + 50=15 + 50 = 65^{\circ}\) and \(m\angle B=x + 100=15+100 = 165^{\circ}\).

Step3: Solve for \(x\) in problem 30

Given \(m\angle A = 6x^{\circ}\) and \(m\angle B=(x + 5)^{\circ}\), we substitute into the equation:
\(6x+(x + 5)=180\).
Combine like - terms: \(7x+5 = 180\).
Subtract 5 from both sides: \(7x=180 - 5 = 175\).
Divide both sides by 7: \(x = 25\).
Then \(m\angle A=6x=6\times25 = 150^{\circ}\) and \(m\angle B=x + 5=25 + 5 = 30^{\circ}\).

Step4: Solve for \(x\) in problem 31

Given \(m\angle A=(2x + 3)^{\circ}\) and \(m\angle B=(3x-223)^{\circ}\), we substitute into the equation:
\((2x + 3)+(3x-223)=180\).
Combine like - terms: \(5x-220 = 180\).
Add 220 to both sides: \(5x=180 + 220=400\).
Divide both sides by 5: \(x = 80\).
Then \(m\angle A=2x + 3=2\times80+3 = 163^{\circ}\) and \(m\angle B=3x-223=3\times80 - 223=17^{\circ}\).

Step5: Solve for \(x\) in problem 32

Given \(m\angle A=(-4x + 40)^{\circ}\) and \(m\angle B=(x + 50)^{\circ}\), we substitute into the equation:
\((-4x + 40)+(x + 50)=180\).
Combine like - terms: \(-3x+90 = 180\).
Subtract 90 from both sides: \(-3x=180 - 90 = 90\).
Divide both sides by \(-3\): \(x=-30\).
Then \(m\angle A=-4x + 40=-4\times(-30)+40=120 + 40 = 160^{\circ}\) and \(m\angle B=x + 50=-30 + 50 = 20^{\circ}\).

Answer:

Problem 29: \(m\angle A = 65^{\circ}\), \(m\angle B = 115^{\circ}\)
Problem 30: \(m\angle A = 150^{\circ}\), \(m\angle B = 30^{\circ}\)
Problem 31: \(m\angle A = 163^{\circ}\), \(m\angle B = 17^{\circ}\)
Problem 32: \(m\angle A = 160^{\circ}\), \(m\angle B = 20^{\circ}\)