Question

2
(11x + 4)°
128°
8x°
m∠fae = 45
m∠acd =

Explanation:

Step1: Find the measure of ∠ABC

Since ∠ABF is a straight angle (180°), and ∠FBC is 128°, we can find ∠ABC by subtracting 128° from 180°.
$$m\angle ABC = 180^{\circ}- 128^{\circ}=52^{\circ}$$

Step2: Use the exterior angle theorem for triangle ABC

The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. For triangle ABC, ∠ACD is an exterior angle, and the two non - adjacent interior angles are ∠ABC and ∠BAC (where ∠BAC = 8x° and ∠ACD=(11x + 4)°). So we have the equation:
$$11x + 4=8x + 52$$
Subtract 8x from both sides:
$$11x-8x+4=8x - 8x+52$$
$$3x + 4=52$$
Subtract 4 from both sides:
$$3x+4 - 4=52 - 4$$
$$3x=48$$
Divide both sides by 3:
$$x = 16$$

Step3: Find \(m\angle FAE\)

∠FAE is a right angle? Wait, no. Wait, ∠FAB and ∠EAB are supplementary? Wait, no. Wait, ∠FAE: Wait, looking at the diagram, AE and AB are perpendicular? Wait, no, ∠FAB and ∠EAB: Wait, actually, ∠FAE is a straight angle? No, wait, the diagram shows that AE is a vertical line and AB is a horizontal line, so ∠FAB and ∠EAB are right angles? Wait, no, wait, the angle at A: ∠FAB and ∠EAB. Wait, actually, ∠FAE is 90°? Wait, no, wait, we found x = 16, and ∠BAC=8x°. Let's check:
Wait, maybe I made a mistake. Wait, ∠FAE: looking at the diagram, AE is a vertical line (downward from A) and AF is a horizontal line (to the right from B, but A is connected to B horizontally). Wait, actually, ∠FAB and ∠EAB: if AB is horizontal and AE is vertical, then ∠FAB and ∠EAB are right angles (90°). Wait, but let's re - examine. Wait, the problem asks for \(m\angle FAE\). Wait, maybe ∠FAE is a straight angle? No, that doesn't make sense. Wait, maybe there is a typo, but according to the diagram, AB is horizontal, AE is vertical, so ∠FAB and ∠EAB are 90°? Wait, no, let's go back. Wait, we found x = 16. Then ∠BAC=8x=8×16 = 128°? No, that can't be. Wait, I must have messed up the exterior angle. Wait, the exterior angle ∠ACD=(11x + 4)°, and the two non - adjacent interior angles are ∠ABC (52°) and ∠BAC (8x°). So the correct formula is \(m\angle ACD=m\angle ABC + m\angle BAC\), so \(11x + 4=52+8x\), which is what I did before, and x = 16. Then ∠BAC=8×16 = 128°? That can't be, because in a triangle, the sum of angles should be less than 180°. Wait, I see my mistake. The exterior angle should be equal to the sum of the two remote interior angles. Wait, ∠ACD is an exterior angle of triangle ABC at C, so the two remote interior angles are ∠ABC and ∠BAC. But if ∠BAC = 8x and ∠ABC=52°, and ∠ACD=11x + 4, then 11x+4=52 + 8x, x = 16. But then ∠BAC=128°, and ∠ABC=52°, then ∠ACB=180-(128 + 52)=0°, which is impossible. So I must have misidentified the angles.

Wait, maybe ∠ABC is not 52°, but the other angle. Wait, ∠FBC is 128°, so ∠ABC=180 - 128=52°, that's correct. Wait, maybe the triangle is not ABC but another triangle. Wait, maybe the angle at A is 8x, and the angle at C is (11x + 4), and the angle at B is 52°, but that gives a degenerate triangle. So I must have made a mistake in the exterior angle. Wait, maybe ∠ACD is an exterior angle at C, and the two remote interior angles are ∠BAC (8x) and ∠ABC (52°), but that's wrong. Wait, maybe the angle at A is not 8x, but the angle between AC and AB is 8x, and the angle between AC and CD is (11x + 4), and AB is parallel to CD? No, the diagram doesn't show that. Wait, maybe it's a right triangle at A. So ∠BAC=90°, then 8x=90, x = 11.25, but that doesn't fit with the exterior angle. Wait, I think I misread the diagram. Let's start over.

Alternative approach:
Looking at the diagram, AB is horizontal, AE is vertical, so ∠FAB and ∠EAB are right angles (90°). So \(m\angle FAE = 90^{\circ}\)? But that seems odd. Wait, the problem has \(m\angle FAE=\) and \(m\angle ACD=\). Wait, maybe the diagram is a right triangle at A, so ∠BAC = 90°, then 8x=90, x = 11.25, but then ∠ACD=11x + 4=11×11.25+4=123.75 + 4=127.75, and ∠ABC=180 - 128=52, and 90+52=142≠127.75. So that's wrong.

Wait, going back to the exterior angle theorem. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So if we consider triangle ABC, with exterior angle at C: ∠ACD, and the two non - adjacent interior angles are ∠ABC and ∠BAC. ∠ABC=180 - 128=52°, ∠BAC=8x°, ∠ACD=(11x + 4)°. So 11x + 4=52+8x→3x=48→x = 16. Then ∠BAC=8×16 = 128°, ∠ABC=52°, then ∠ACB=180-(128 + 52)=0°, which is impossible. So there must be a mistake in the identification of the exterior angle.

Wait, maybe the exterior angle is at B. Wait, no. Wait, the angle at B: ∠FBC=128°, which is an exterior angle of triangle ABC at B. Then the two non - adjacent interior angles would be ∠BAC and ∠ACB. So 128=8x + ∠ACB. And ∠ACD=(11x + 4), and ∠ACD and ∠ACB are supplementary (since D - C - B is a straight line? Wait, D, C, B: no, D, C are on one line, and C, B are on another line. Wait, the diagram: D---C--- (then C to B, and C to A, A to B, A to E). So DC is a line, CA is a line, AB is a line, BF is a line, AE is a line.

Wait, maybe ∠ACD and ∠ACB are supplementary, so \(m\angle ACD + m\angle ACB=180^{\circ}\). And in triangle ABC, \(m\angle BAC + m\angle ABC + m\angle ACB=180^{\circ}\). We know \(m\angle ABC = 180 - 128 = 52^{\circ}\), \(m\angle BAC=8x^{\circ}\), \(m\angle ACD=(11x + 4)^{\circ}\), so \(m\angle ACB=180-(11x + 4)^{\circ}\). Then:

\(8x+52+(180-(11x + 4))=180\)

Simplify:

\(8x + 52+180-11x - 4=180\)

\(- 3x+228=180\)

\(-3x=180 - 228=-48\)

\(x = 16\)

Ah, so even though ∠BAC=8×16 = 128°, and ∠ABC=52°, and ∠ACB=180-(11×16 + 4)=180 - 180 = 0°, which is impossible. So there must be a misinterpretation of the diagram.

Wait, maybe the angle at A is not 8x, but the angle between AE and AC is 8x? No, the diagram says 8x° at A, between AB and AC.

Wait, maybe the problem has a typo, but assuming our calculation of x = 16 is correct (from the exterior angle equation which led to x = 16, even though the triangle seems degenerate, maybe it's a different configuration).

Now, \(m\angle FAE\): looking at the diagram, AE and AF are perpendicular? So \(m\angle FAE = 90^{\circ}\)? But that seems odd. Wait, maybe the diagram is such that AB is horizontal, AE is vertical, so ∠FAB and ∠EAB are right angles, so ∠FAE is a straight angle? No, that can't be. Wait, maybe the question is mislabeled, and \(m\angle FAB = 90^{\circ}\), so \(m\angle FAE = 90^{\circ}\).

For \(m\angle ACD\): substitute x = 16 into \(11x + 4\):

\(11×16+4=176 + 4=180^{\circ}\)? No, that can't be. Wait, this is a problem. There must be a mistake in my angle identification.

Wait, let's start over. Let's assume that ∠FAE is 90° (since AB is horizontal and AE is vertical, forming a right angle). Then, in triangle ABC, ∠BAC=90°, ∠ABC=52°, so ∠ACB=38°. Then ∠ACD=180 - 38 = 142°. And 11x + 4=142→11x=138→x≈12.55, which doesn't match 8x=90→x = 11.25. So there is a contradiction.

Wait, maybe the exterior angle is ∠ACD, and the two non - adjacent interior angles are ∠BAC (8x) and ∠ABC (180 - 128 = 52). So 11x+4=8x + 52→3x=48→x = 16. Then ∠ACD=11×16+4=180°, which means that points D, C, and B are colinear? But in the diagram, C is connected to B and D. So if ∠ACD=180°, then D, C, B are colinear. So the triangle is degenerate, with C on the line DB. So in that case, ∠FAE: if AB is horizontal and AE is vertical, then ∠FAE=90°.

So, despite the degenerate triangle, we'll go with x = 16.

Step4: Calculate \(m\angle FAE\)

If AB is horizontal and AE is vertical, then ∠FAE is a right angle, so \(m\angle FAE = 90^{\circ}\). Wait, but 8x=8×16 = 128°, which would mean ∠BAC=128°, which is more than 90°, so AB and AE can't be perpendicular. So maybe ∠FAE is a straight angle? No, that would be 180°. I think there is a mistake in the problem or my interpretation. But following the exterior angle theorem calculation:

We have x = 16.

\(m\angle FAE\): Wait, maybe the angle at A is 8x, and since AB and AE are such that ∠FAE is 180° - 8x? No, that doesn't make sense. Wait, the problem says "m∠FAE = " and "m∠ACD = ". Let's proceed with the values we got from x = 16.

\(m\angle ACD=11x + 4=11×16+4=180^{\circ}\)

\(m\angle FAE\): If we consider that ∠FAB is 180° - 8x (since ∠BAC=8x and ∠FAE is supplementary to ∠BAC? No, that doesn't make sense. Wait, maybe the diagram is different. Maybe ∠FAE is equal to ∠BAC? No.

Wait, maybe the original problem has a different diagram, but based on the given equations, we have:

From the exterior angle theorem: 11x + 4=8x+(180 - 128)

11x + 4=8x + 52

3x=48

x = 16

Then \(m\angle ACD=11×16 + 4=180^{\circ}\)

\(m\angle FAE\): If AB and AE are perpendicular, then \(m\angle FAE = 90^{\circ}\), but 8x=128°, which is not 90°. So there is an error. But maybe the problem assumes that ∠FAE is 90°, and we proceed.

Answer:

\(m\angle FAE=\boldsymbol{90^{\circ}}\) (assuming AB and AE are perpendicular)
\(m\angle ACD=\boldsymbol{180^{\circ}}\) (from x = 16, 11x+4=180)

Wait, but this is a degenerate case. Maybe I made a mistake in the exterior angle identification. Let's try another way. Suppose that ∠ACD is equal to ∠BAC + ∠ABC, and ∠ABC=128°? No, ∠ABC is adjacent to 128°, so ∠ABC=52°. If ∠ACD=∠BAC + ∠ABC, and ∠BAC=8x, ∠ABC=52, ∠ACD=11x + 4. So 11x + 4=8x + 52→x = 16. Then ∠ACD=180°, which means that D, C, B are colinear. So the triangle is degenerate, with C on DB. In that case, ∠FAE: if AB is horizontal and AE is vertical, then ∠FAE=90°, and ∠BAC=8x=128°, which would mean that AC is a line from A to C, where C is on DB, so ∠BAC=128°, which is possible in a degenerate triangle.

So, despite the degeneracy, the answers are \(m\angle FAE = 90^{\circ}\) (assuming right angle at A) and \(m\angle ACD=180^{\circ}\) (since D, C, B are colinear).