Question

1. diagram with intersecting lines, angles (3x)°, 49°, (7x - 23)°, and lines l, m with angle (11y - 1)°. 12. diagram with a right triangle (right angle) between lines l and m, angles (7r - 20)°, (5r - 38)°, (3r - 4)°.

Explanation:

Problem 11:

Step 1: Find x using vertical angles

Vertical angles are equal, so \(7x - 23 = 49 + 3x\)
Subtract \(3x\) from both sides: \(4x - 23 = 49\)
Add 23 to both sides: \(4x = 72\)
Divide by 4: \(x = 18\)

Step 2: Find y using parallel lines (if l || m, corresponding angles are equal)

Assuming lines \(l\) and \(m\) are parallel, then \(11y - 1 = 49 + 3x\) (corresponding angles)
We know \(x = 18\), so \(3x = 54\)
Then \(11y - 1 = 49 + 54 = 103\)
Add 1 to both sides: \(11y = 104\) (Wait, maybe alternate interior angles? Let's check again. If \(l\) and \(m\) are parallel, then \(11y - 1\) and \(7x - 23\) might be equal? Wait, maybe the angle \(3x\) and \(11y - 1\) are equal as alternate interior angles. Wait, let's re - examine the diagram. If two lines are crossed by a transversal, and \(l\) and \(m\) are parallel, then alternate interior angles are equal. So \(3x=11y - 1\)
We know \(x = 18\), so \(3\times18=11y - 1\)
\(54 = 11y-1\)
Add 1 to both sides: \(11y=55\)
Divide by 11: \(y = 5\)

Problem 12:

Step 1: Use the fact that the sum of angles in a triangle is 180°, and there is a right angle (90°)

So \((7y - 20)+(5y - 38)+(3y - 4)=90\) (since it's a right - angled triangle, the two non - right angles and the right angle? Wait, no. Wait, the diagram shows a right angle between the two lines, so the triangle is a right - triangle with the right angle between the two parallel lines? Wait, the sum of the three angles in the triangle: \((7y - 20)+(5y - 38)+(3y - 4)=180 - 90?\) No, wait, the three angles of the triangle: one is \((7y - 20)\), one is \((5y - 38)\), and one is \((3y - 4)\), and since it's a triangle formed by two parallel lines and a transversal, and there is a right angle, so the sum of the three angles should be 180°, and one of them is 90°? Wait, no. Wait, the angle with the right - angle symbol is 90°, so the other two angles \((7y - 20)\) and \((5y - 38)\) and \((3y - 4)\)? Wait, no, the right angle is between the two lines, so the triangle has angles \((7y - 20)\), \((5y - 38)\) and \((3y - 4)\), and since it's a triangle, \((7y - 20)+(5y - 38)+(3y - 4)=180\)? Wait, no, the right angle is 90°, so \((7y - 20)+(5y - 38)+(3y - 4)=90\)? No, that can't be. Wait, maybe the two angles \((7y - 20)\) and \((5y - 38)\) are the non - right angles, and \((3y - 4)\) is part of the right angle? Wait, no. Let's re - think. The sum of the interior angles of a triangle is 180°. If one angle is 90° (the right angle), then the sum of the other two angles is 90°. Wait, maybe \((7y - 20)+(5y - 38)=90\) and \((3y - 4)\) is equal to something? No, the diagram shows a right angle at the intersection of the two lines, so the triangle has angles: \((7y - 20)\), \((5y - 38)\) and \((3y - 4)\), and since it's a triangle, \((7y - 20)+(5y - 38)+(3y - 4)=180\)
Combine like terms: \(7y+5y + 3y-20 - 38 - 4=180\)
\(15y-62 = 180\)
Add 62 to both sides: \(15y=242\) (This seems wrong. Wait, maybe the right angle is such that \((3y - 4)+(7y - 20)+(5y - 38)=180\), but the right angle is outside? No, the diagram has a right - angle symbol between the two lines, so the triangle is a right - triangle with the right angle at the vertex on the line \(m\). So the two angles \((7y - 20)\) and \((5y - 38)\) are the acute angles, and \((3y - 4)\) is adjacent to the right angle? Wait, no. Let's try again. If the two lines are parallel, and we have a transversal forming a right - angled triangle, then the sum of the three angles in the triangle is 180°, and one angle is 90°. So:

\((7y - 20)+(5y - 38)+(3y - 4)=180\)

\(15y-62 = 180\)

\(15y=242\)

\(y=\frac{242}{15}\approx16.13\) (This seems incorrect. Maybe the right angle is 90°, so \((7y - 20)+(5y - 38)=90\) (since the third angle is 90°? No, the right - angle symbol is at the vertex where \((3y - 4)\) is. So \((3y - 4) = 90\)? No, \((3y - 4)\) is an angle in the triangle. Wait, maybe the two lines are parallel, so the alternate interior angles or something. Wait, perhaps the correct equation is \((7y - 20)+(5y - 38)+(3y - 4)=180\), but let's check the arithmetic again:

\(7y+5y + 3y=15y\)

\(-20-38 - 4=-62\)

So \(15y=180 + 62=242\), \(y=\frac{242}{15}\approx16.13\). But this seems odd. Maybe there is a mistake in my understanding. Alternatively, if the right angle is 90°, then \((3y - 4)+(7y - 20)+(5y - 38)=180\), and \((3y - 4)\) is not the right angle. Wait, maybe the diagram is a right - angled triangle with the right angle between the two parallel lines, so the angle between the two parallel lines is 90°, and the triangle has angles \((7y - 20)\), \((5y - 38)\) and \((3y - 4)\), and since the sum of a…

Answer:

s:

Problem 11:

\(x = 18\), \(y = 5\)

Problem 12:

\(y=\frac{242}{15}\approx16.13\) (or if there is a misinterpretation, let's re - check. Wait, maybe the right angle is 90°, so \((7y - 20)+(5y - 38)=90\) (since the third angle \((3y - 4)\) is equal to the right angle? No, that doesn't make sense. Alternatively, maybe the two angles \((7y - 20)\) and \((5y - 38)\) are equal to \((3y - 4)\) in some way. Wait, perhaps the lines are parallel, so alternate interior angles: \((7y - 20)=(3y - 4)\) and \((5y - 38)\) is something else. Let's try that:

\(7y-20 = 3y - 4\)

\(7y-3y=-4 + 20\)

\(4y = 16\)

\(y = 4\)

Then check \((5y - 38)=20 - 38=-18\), which is impossible (angle can't be negative). So my initial approach was wrong. Let's try again. The triangle has a right angle, so the sum of the two non - right angles is 90°. So \((7y - 20)+(5y - 38)=90\)

\(12y-58 = 90\)

\(12y=148\)

\(y=\frac{148}{12}=\frac{37}{3}\approx12.33\)

And \((3y - 4)\) is the right angle? \(3\times\frac{37}{3}-4=37 - 4 = 33
eq90\). No. Alternatively, \((3y - 4)\) is one of the acute angles, and \((7y - 20)+(5y - 38)+(3y - 4)=180\), and the right angle is 90°, so \((7y - 20)+(5y - 38)=90\), then \((3y - 4)=90\), \(3y=94\), \(y=\frac{94}{3}\approx31.33\), then \((7y - 20)+(5y - 38)=12y - 58=12\times\frac{94}{3}-58 = 376-58 = 318
eq90\). I think there is a misinterpretation of the diagram. Assuming that the triangle is a right - triangle with the right angle, and the three angles are \((7y - 20)\), \((5y - 38)\) and \((3y - 4)\), and the sum is 180°, then \(y=\frac{242}{15}\approx16.13\) is the answer.