Question

if a transversal crosses parallel lines then alternate exterior angles are congruent. line t crosses parallel line m and n.
a ∠1 is congruent to ∠7
b ∠1 is congruent to ∠5
c ∠4 is congruent to ∠8
d ∠4 is congruent to ∠6

1. what is the value of x?

a 16
b 20
c 21
d 28

Explanation:

Step1: Analyze the angles on a straight line and right angle

The sum of angles on a straight line is \(180^\circ\), and there is a right angle (\(90^\circ\)) in the diagram. So, we have the equation: \((5y + 10)^\circ + y^\circ + 90^\circ + (3x + 7)^\circ = 180^\circ\)? Wait, no, actually, the vertical angles and linear pairs. Wait, the angle \((5y + 10)^\circ\) and its vertical angle, and the right angle. Wait, let's re - examine. The angle \((5y + 10)^\circ\) and the angle adjacent to it (with \(y^\circ\) and the right angle) form a linear pair? Wait, no, the straight line has a right angle, so the sum of \((5y + 10)^\circ\), \(y^\circ\) and \((3x + 7)^\circ\) plus the right angle? Wait, no, the right angle is \(90^\circ\), and the three angles \((5y + 10)^\circ\), \(y^\circ\) and \((3x + 7)^\circ\) and the right angle? Wait, no, the straight line is \(180^\circ\), so \((5y + 10)^\circ + y^\circ+ 90^\circ+(3x + 7)^\circ = 180^\circ\)? No, that's not right. Wait, actually, the angle \((5y + 10)^\circ\) and the angle \((3x + 7)^\circ\) are vertical angles? No, the angle \((5y + 10)^\circ\) and the angle opposite to it (the one with \((3x + 7)^\circ\))? Wait, no, the angle \((5y + 10)^\circ\) and \(y^\circ\) and the right angle: since the right angle is \(90^\circ\), and \((5y + 10)^\circ\) and \(y^\circ\) are complementary to the right angle? Wait, no, let's look at the linear pair. The angle \((5y + 10)^\circ\) and the angle formed by \(y^\circ\) and the right angle: \((5y + 10)^\circ + y^\circ+ 90^\circ= 180^\circ\) (because they are on a straight line). So first, solve for \(y\):
\(5y + 10 + y+ 90 = 180\)
\(6y+100 = 180\)
\(6y=180 - 100\)
\(6y = 80\)? No, that can't be. Wait, maybe I made a mistake. Wait, the angle \((5y + 10)^\circ\) and the angle \(y^\circ\) and the right angle: actually, the angle \((5y + 10)^\circ\) and \(y^\circ\) are such that \((5y + 10)^\circ\) and \(y^\circ\) and the right angle: no, the correct approach is that the angle \((5y + 10)^\circ\) and the angle \(y^\circ\) are complementary to the right angle? Wait, no, let's use the fact that the angle \((5y + 10)^\circ\) and its vertical angle, and the angle \((3x + 7)^\circ\) and the right angle. Wait, another way: the angle \((5y + 10)^\circ\) and \(y^\circ\) are adjacent to the right angle, so \((5y + 10)+y + 90=180\) (since they are on a straight line).
So, \(6y+100 = 180\)
\(6y=180 - 100\)
\(6y = 80\)? No, that gives a non - integer, which is wrong. Wait, maybe the angle \((5y + 10)^\circ\) and \(y^\circ\) are vertical angles? No, that's not. Wait, maybe the angle \((5y + 10)^\circ\) and the angle \((3x + 7)^\circ\) are vertical angles? No, the angle \((5y + 10)^\circ\) and \(y^\circ\) are related as \((5y + 10)=y + 90\)? Wait, because the angle \((5y + 10)^\circ\) and the angle formed by \(y^\circ\) and the right angle are vertical angles? Wait, if that's the case, then \(5y+10=y + 90\)
\(5y - y=90 - 10\)
\(4y = 80\)
\(y = 20\)

Step2: Substitute \(y\) to find \(x\)

Now that we have \(y = 20\), the angle \((5y + 10)^\circ=(5\times20 + 10)^\circ=110^\circ\). The angle \((3x + 7)^\circ\) and \((5y + 10)^\circ\) are supplementary? Wait, no, the angle \((3x + 7)^\circ\) and the angle \((5y + 10)^\circ\) are vertical angles? No, wait, the straight line: the angle \((5y + 10)^\circ\) and \((3x + 7)^\circ\) and the right angle? Wait, no, when \(y = 20\), the angle \(y^\circ = 20^\circ\), and the right angle is \(90^\circ\), so the angle \((3x + 7)^\circ\) should be equal to \(y^\circ\)? No, wait, the angle \((3x + 7)^\circ\) and \(y^\circ\) are vertical angles? Wait, no, let's use the fact that the sum of angles on a straight line is \(180^\circ\). We know that \((5y + 10)^\circ+(3x + 7)^\circ= 180^\circ - 90^\circ=90^\circ\)? No, that's not right. Wait, when \(y = 20\), \((5y + 10)=110\), and the angle \((3x + 7)\) and \((5y + 10)\) are supplementary? Wait, no, the straight line has a right angle, so the sum of \((5y + 10)\), \(y\) and \((3x + 7)\) plus the right angle is \(180\)? No, we found \(y = 20\) from \(5y + 10=y + 90\) (since they are vertical angles with the right angle - related angle). Now, the angle \((3x + 7)^\circ\) and \(y^\circ\) are equal? Wait, no, the angle \((3x + 7)^\circ\) and \(y^\circ\) are vertical angles? Wait, when \(y = 20\), we can find \((3x + 7)^\circ\) such that \((3x + 7)+(5y + 10)=180\)? No, \((5y + 10)=110\), so \(3x + 7=180 - 110 = 70\)? No, that's not. Wait, no, the angle \((3x + 7)^\circ\) and \(y^\circ\) are equal because they are vertical angles? Wait, \(y = 20\), so \(3x+7 = 20\)? No, that gives \(3x=13\), which is not an option. Wait, I think I made a mistake in the first step. Let's start over.
The correct approach: The angle \((5y + 10)^\circ\) and the angle \(y^\circ\) and the right angle (\(90^\circ\)) and the angle \((3x + 7)^\circ\) are on a straight line, so their sum is \(180^\circ\). So:
\((5y + 10)+y + 90+(3x + 7)=180\)
\(6y+3x+107 = 180\)
\(6y+3x=180 - 107\)
\(6y+3x = 73\) (This can't be right as \(y\) and \(x\) should be integers from the options). Wait, another way: The angle \((5y + 10)^\circ\) and the angle \((3x + 7)^\circ\) are vertical angles? No, the angle \((5y + 10)^\circ\) and the angle \(y^\circ\) are complementary to the right angle? Wait, the right angle is \(90^\circ\), so \(y+(5y + 10)=90\) (because they are complementary to the right angle, forming a right angle together). So:
\(y+5y + 10=90\)
\(6y=90 - 10\)
\(6y = 80\) (No, still wrong). Wait, maybe the angle \((5y + 10)^\circ\) and the angle \((3x + 7)^\circ\) are equal because they are vertical angles, and \(y^\circ\) and the right angle and \((5y + 10)^\circ\) form a linear pair. Wait, let's look at the diagram again. The angle \((5y + 10)^\circ\) and the angle \((3x + 7)^\circ\) are vertical angles? No, the angle \((5y + 10)^\circ\) and \(y^\circ\) are adjacent, and the right angle is between them? Wait, I think the correct equation is that the angle \((5y + 10)^\circ\) and \(y^\circ\) are such that \((5y + 10)+y=90\) (since they are complementary to the right angle), so:
\(6y+10 = 90\)
\(6y=80\) (No, not integer). Wait, the problem must have the angle \((5y + 10)^\circ\) and \((3x + 7)^\circ\) as vertical angles, and \(y^\circ\) and the right angle: Wait, when we look at the options, let's substitute \(x\) values.
Option A: \(x = 16\), then \((3x + 7)=3\times16+7 = 55\)
Option B: \(x = 20\), \((3x + 7)=67\)
Option C: \(x = 21\), \((3x + 7)=70\)
Option D: \(x = 28\), \((3x + 7)=91\)
Now, let's assume that the angle \((5y + 10)^\circ\) and \((3x + 7)^\circ\) are supplementary to the angle \(y^\circ\) and the right angle. Wait, if we consider that the angle \((5y + 10)^\circ\) and \(y^\circ\) are such that \(5y + 10=y + 90\) (as vertical angles with the right - angle - included angle), then \(y = 20\), as before. Then the angle \((3x + 7)^\circ\) should be equal to \(y^\circ = 20^\circ\)? No, that's not. Wait, when \(y = 20\), the angle \((5y + 10)^\circ=110^\circ\), and the angle \((3x + 7)^\circ\) should be \(70^\circ\) (since \(110+70 = 180\)). So \(3x+7 = 70\)
\(3x=70 - 7\)
\(3x = 63\)
\(x = 21\)

Answer:

C. 21