find the values of x and y.
(2x)°
128°
x =
y =

Question

Accepted Answer

# Explanation:
## Step1: Find x using angle sum
The sum of angles around a point on a straight line (or semicircle - related here) and the right angle. First, note that the angle between $FB$ and $FC$ is $90^\circ$ (right angle), and the red arc is $128^\circ$, and the angle $(2x)^\circ$ and the angle from $FB$ to $FA$. Wait, actually, the sum of the angle $128^\circ$, $(2x)^\circ$, and the right angle? Wait no, looking at the diagram, the angle from $FA$ to $FB$ is $2x$, from $FB$ to $FC$ is $128^\circ$? Wait no, the right angle is between $FB$ and $FC$? Wait, no, the right angle is between $FC$ and $FE$? Wait, no, the diagram shows $FC$ is horizontal right, $FB$ is vertical up, so angle between $FB$ and $FC$ is $90^\circ$? Wait, no, the red arc is from $FA$ to $FC$? Wait, no, the red arc is from $A$ to $C$ passing through $B$, with $128^\circ$. Wait, actually, the angle around point $F$: the angle between $FA$ and $FB$ is $2x$, between $FB$ and $FC$ is $128^\circ$, and between $FC$ and $FE$ is $90^\circ$? Wait, no, let's re - examine. The key is that the sum of the angle $128^\circ$, $2x^\circ$, and the right angle? Wait, no, actually, the angle from $FA$ to $FC$ should be $180^\circ$? Wait, no, the straight line? Wait, maybe the angle between $FA$, $FB$, and $FC$: the angle $2x + 128^\circ$ and the right angle? Wait, no, I think I made a mistake. Let's look again. The angle between $FB$ (vertical up) and $FC$ (horizontal right) is $90^\circ$, but the red arc is $128^\circ$ from $A$ to $C$ through $B$. So the sum of $2x$, $128^\circ$, and the right angle? No, wait, the angle from $FA$ to $FB$ is $2x$, from $FB$ to $FC$ is $128^\circ$, but the angle between $FB$ and $FC$ should be such that $2x+128^\circ = 180^\circ - 90^\circ$? No, wait, no. Wait, the total angle around a point for a straight line (semicircle) is $180^\circ$? Wait, no, the angle between $FA$, $FB$, and $FC$: actually, the angle from $FA$ to $FC$ is $180^\circ$ (since $FA$ and $FC$ are on a straight line? No, $FA$ is a line, $FC$ is another line. Wait, maybe the angle between $FA$ and $FB$ is $2x$, between $FB$ and $FC$ is $128^\circ$, and since the angle between $FB$ and $FC$ plus $2x$ plus the right angle? No, I think the correct approach is: the sum of the angle $128^\circ$, $2x^\circ$, and the right angle (90°) should be... Wait, no, let's use the fact that the angle on a straight line (or the sum of angles around a point for a semicircle - related). Wait, actually, the angle between $FA$ and $FC$ is $180^\circ$ (since $FA$ and $FC$ are on a straight line? No, $FA$ is going up - left, $FC$ is going right. Wait, maybe the angle from $FA$ to $FB$ is $2x$, from $FB$ to $FC$ is $128^\circ$, and the angle between $FB$ and $FC$ is $128^\circ$, and the angle between $FB$ and $FA$ is $2x$, and the angle between $FA$ and $FC$ is $2x + 128^\circ$, but there is a right angle? Wait, no, the right angle is between $FC$ and $FE$? Wait, I think I messed up. Let's start over.

Looking at the diagram, $FC$ is horizontal right, $FB$ is vertical up, so angle between $FB$ and $FC$ is $90^\circ$? No, the red arc is $128^\circ$ from $A$ to $C$ passing through $B$. So the angle from $FA$ to $FB$ is $2x$, from $FB$ to $FC$ is $128^\circ$, and the angle from $FC$ to $FE$ is $90^\circ$? No, the right angle is between $FC$ and $FE$? Wait, no, the right angle symbol is between $FC$ and $FE$? No, the right angle is between $FC$ and $FB$? Wait, the diagram shows a right angle between $FC$ (horizontal right) and $FE$ (vertical down)? No, $FB$ is vertical up, $FC$ is horizontal right, so angle between $FB$ and $FC$ is $90^\circ$. Then the angle from $FA$ to $FB$ is $2x$, from $FB$ to $FC$ is $128^\circ$, but that can't be because $90^\circ+128^\circ>180^\circ$. Wait, I think the correct way is: the sum of the angle $2x$, $128^\circ$, and the right angle (90°) is wrong. Wait, actually, the angle around point $F$: the angle from $FA$ to $FC$ is $180^\circ$ (since $FA$ and $FC$ are on a straight line? No, $FA$ is a line, $FC$ is another line. Wait, maybe the angle between $FA$ and $FB$ is $2x$, between $FB$ and $FC$ is $128^\circ$, and the angle between $FC$ and $FD$ is $(y + 5)^\circ$, and between $FD$ and $FE$ is... Wait, no, let's use the fact that vertical angles or linear pairs.

Wait, the key is that the angle $2x$ and the angle $(y + 5)^\circ$ are vertical angles? No, wait, the angle between $FA$ and $FB$ is $2x$, and the angle between $FD$ and $FE$ is equal to $2x$ (vertical angles). Also, the angle between $FB$ and $FC$ is $128^\circ$, and the angle between $FC$ and $FD$ is $(y + 5)^\circ$, and the angle between $FD$ and $FE$ is $90^\circ - (y + 5)^\circ$? No, this is getting confusing. Let's try a different approach.

First, find $x$: The sum of the angle $2x$, $128^\circ$, and the right angle (90°) should be... Wait, no, the angle from $FA$ to $FC$ is $180^\circ$ (since $FA$ and $FC$ are on a straight line? No, $FA$ is going up - left, $FC$ is going right. Wait, maybe the angle between $FA$, $FB$, and $FC$: the angle $2x + 128^\circ= 180^\circ$? No, because there is a right angle. Wait, I think the correct equation is $2x+128^\circ = 180^\circ$? No, that would be if they are supplementary, but there is a right angle. Wait, no, looking at the diagram, the angle between $FB$ (vertical up) and $FC$ (horizontal right) is $90^\circ$, but the red arc is $128^\circ$ from $A$ to $C$ through $B$. So the angle from $A$ to $B$ is $2x$, from $B$ to $C$ is $128^\circ$, and from $A$ to $C$ should be $180^\circ$ (since it's a semicircle? Wait, the red arc is a semicircle? No, a semicircle is $180^\circ$. So $2x + 128^\circ= 180^\circ$? Wait, that would mean $2x=180 - 128 = 52$, so $x = 26$. Wait, that makes sense. Let's check: if $x = 26$, then $2x=52^\circ$, and $52 + 128=180^\circ$, which is a straight line (semicircle), so that works.

## Step2: Find y using vertical angles
Now, the angle $2x = 52^\circ$ and the angle $(y + 5)^\circ$ are vertical angles? Wait, no, the angle between $FD$ and $FE$ is equal to $2x$? Wait, no, the right angle is between $FC$ and $FE$ (vertical down), so the angle between $FC$ (horizontal right) and $FD$ is $(y + 5)^\circ$, and the angle between $FD$ and $FE$ (vertical down) is $90^\circ-(y + 5)^\circ$. But also, the angle between $FA$ and $FB$ is $2x = 52^\circ$, and the angle between $FD$ and $FE$ is equal to $2x$ (vertical angles). So $90-(y + 5)=2x$. Wait, we know $2x = 52$, so $90-(y + 5)=52$. Then $y + 5=90 - 52=38$, so $y=38 - 5 = 33$. Wait, let's verify:

If $x = 26$, then $2x = 52^\circ$. The angle between $FD$ and $FE$ should be equal to $52^\circ$ (vertical angles with $2x$). The angle between $FC$ (horizontal right) and $FE$ (vertical down) is $90^\circ$, so the angle between $FC$ and $FD$ is $90 - 52=38^\circ$, which is $(y + 5)^\circ$. So $y + 5=38$, so $y = 33$.

## Step1 (correct) for x:
The angle $2x$ and the $128^\circ$ angle are supplementary (sum to $180^\circ$) because they form a straight line (semicircle - like). So:
$2x+128 = 180$
Subtract 128 from both sides:
$2x=180 - 128=52$
Divide by 2:
$x = 26$

## Step2 (correct) for y:
The angle $2x = 52^\circ$ and the angle $(y + 5)^\circ$ are related to the right angle. The angle between $FC$ (horizontal) and $FE$ (vertical) is $90^\circ$. The angle between $FD$ and $FE$ is equal to $2x$ (vertical angles), so the angle between $FC$ and $FD$ is $90 - 2x$. But this angle is also $(y + 5)^\circ$. So:
$y + 5=90 - 2x$
We know $x = 26$, so $2x = 52$
$y+5=90 - 52 = 38$
Subtract 5:
$y=38 - 5=33$

# Answer:
$x = 26$, $y = 33$

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