Question

1. find the value of x in the diagrams below.

(3x - 1)°
71°

1. (2x)°

(4x - 6)°

1. ∠a and ∠b are vertical angles. if m∠a=(5x - 28)° and m∠b=(3x + 4)°, find the measure of each angle.
2. ∠p and ∠q form a linear pair. if m∠p=(7x - 4)° and m∠q=(2x - 5)°, find the measure of each angle.

Explanation:

Step1: Recall vertical - angles property

Vertical angles are equal. In problem 1, we set up the equation \(3x - 1=71\).

Step2: Solve for \(x\) in problem 1

Add 1 to both sides: \(3x=71 + 1=72\). Then divide both sides by 3: \(x = \frac{72}{3}=24\).

Step3: Recall vertical - angles property for problem 2

Set up the equation \(2x=4x - 6\).

Step4: Solve for \(x\) in problem 2

Subtract \(2x\) from both sides: \(0=4x-2x - 6\), so \(2x-6 = 0\). Add 6 to both sides: \(2x=6\), then \(x = 3\).

Step5: Recall vertical - angles property for problem 3

Set up the equation \(5x-28=3x + 4\).

Step6: Solve for \(x\) in problem 3

Subtract \(3x\) from both sides: \(5x-3x-28=4\), so \(2x-28 = 4\). Add 28 to both sides: \(2x=4 + 28=32\). Divide both sides by 2: \(x = 16\). Then find \(m\angle A=5x-28=5\times16-28=80 - 28 = 52^{\circ}\), \(m\angle B=3x + 4=3\times16+4=48 + 4=52^{\circ}\).

Step7: Recall linear - pair property for problem 4

Since \(\angle P\) and \(\angle Q\) form a linear pair, \(m\angle P+m\angle Q = 180^{\circ}\). So \((7x-4)+(2x-5)=180\).

Step8: Simplify the left - hand side of the equation in problem 4

Combine like terms: \(7x+2x-4 - 5=180\), \(9x-9 = 180\).

Step9: Solve for \(x\) in problem 4

Add 9 to both sides: \(9x=180 + 9=189\). Divide both sides by 9: \(x = 21\). Then \(m\angle P=7x-4=7\times21-4=147 - 4=143^{\circ}\), \(m\angle Q=2x-5=2\times21-5=42 - 5 = 37^{\circ}\).

Answer:

1. \(x = 24\)
2. \(x = 3\)
3. \(m\angle A = 52^{\circ}\), \(m\angle B = 52^{\circ}\)
4. \(m\angle P=143^{\circ}\), \(m\angle Q = 37^{\circ}\)