Question

the angle measurements in the diagram are represented by the following expressions.
∠a = 7x + 40° ∠b = 3x + 112°
diagram
solve for x and then find the measure of ∠a:
∠a = □°

Explanation:

Step1: Identify angle relationship

From the diagram (parallel lines, so ∠A and ∠B are equal, alternate - interior or corresponding angles). So set \(7x + 40^{\circ}=3x + 112^{\circ}\).

Step2: Solve for x

Subtract \(3x\) from both sides: \(7x-3x + 40^{\circ}=3x-3x + 112^{\circ}\), which simplifies to \(4x+40^{\circ}=112^{\circ}\).
Subtract \(40^{\circ}\) from both sides: \(4x+40^{\circ}-40^{\circ}=112^{\circ}-40^{\circ}\), so \(4x = 72^{\circ}\).
Divide both sides by 4: \(x=\frac{72^{\circ}}{4}=18^{\circ}\).

Step3: Find measure of ∠A

Substitute \(x = 18^{\circ}\) into \(∠A=7x + 40^{\circ}\).
\(∠A=7\times18^{\circ}+40^{\circ}=126^{\circ}+40^{\circ}=166^{\circ}\). Wait, no, wait. Wait, if the lines are parallel, maybe ∠A and ∠B are supplementary? Wait, let's re - examine. If the two angles are on the same side of the transversal, they are supplementary. Let's check the diagram again. The angles ∠A and ∠B, if the lines are parallel, and the transversal cuts them, maybe they are supplementary. So \(7x + 40+3x + 112 = 180\).
Combine like terms: \(10x+152 = 180\).
Subtract 152: \(10x=180 - 152=28\).
Then \(x = 2.8\). Wait, that's a mistake. Wait, the diagram: ∠A and ∠B, if the two lines are parallel, and the transversal is a line that makes ∠A and ∠B same - side interior angles, then they are supplementary. Let's start over.

Step1: Correct angle relationship

Since the lines are parallel, ∠A and ∠B are same - side interior angles, so they are supplementary. So \(∠A+∠B = 180^{\circ}\).
So \((7x + 40^{\circ})+(3x + 112^{\circ})=180^{\circ}\).

Step2: Solve for x

Combine like terms: \(10x+152^{\circ}=180^{\circ}\).
Subtract \(152^{\circ}\) from both sides: \(10x=180^{\circ}-152^{\circ}=28^{\circ}\).
Divide by 10: \(x = 2.8^{\circ}\)? No, that can't be. Wait, maybe the angles are equal. Wait, maybe the diagram is such that ∠A and ∠B are alternate - exterior angles. Let's check the original problem again. The user's diagram: two parallel lines, a transversal, ∠A and ∠B. Maybe I misjudged the relationship. Let's assume that ∠A and ∠B are equal (corresponding angles). So \(7x + 40=3x + 112\).
Subtract 3x: \(4x+40 = 112\).
Subtract 40: \(4x=72\), \(x = 18\). Then ∠A=718 + 40=126 + 40=166, ∠B=318+112 = 54+112 = 166. Oh, so they are equal. So the lines are parallel, and the angles are equal (maybe alternate - interior or corresponding). So that works. So my first thought was correct. So \(x = 18\), ∠A=718 + 40=166? Wait, 718 is 126, 126 + 40 is 166. And ∠B=3*18+112 = 54 + 112=166. So they are equal, so the angle relationship is that they are equal (corresponding angles or alternate - interior angles). So the correct steps are:

Step1: Set angles equal

Since the lines are parallel, ∠A = ∠B (corresponding or alternate - interior angles). So \(7x + 40=3x + 112\).

Step2: Solve for x

Subtract 3x from both sides: \(4x+40 = 112\).
Subtract 40 from both sides: \(4x=72\).
Divide by 4: \(x = 18\).

Step3: Calculate ∠A

Substitute x = 18 into \(∠A=7x + 40\): \(∠A=7\times18+40=126 + 40=166\).

Answer:

\(166\)