Your student will be angling for an A after this missing angle ... - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

Problem 1

Goal: Find angles $A$ and $B$.
* Angle A: This is a right triangle (indicated by the square symbol), which means one angle is $90^\circ$. The sum of angles in any triangle is $180^\circ$.
* Calculation: $180^\circ - 90^\circ - 45^\circ = 45^\circ$. So, $A = 45^\circ$.
* Angle B: Angles on a straight line add up to $180^\circ$. Angle $B$ and the $45^\circ$ angle form a straight line.
* Calculation: $180^\circ - 45^\circ = 135^\circ$. So, $B = 135^\circ$.

Problem 2

Goal: Find angles $A$, $B$, and $C$.
* Angle A: Vertical angles (angles opposite each other where lines cross) are equal. Angle $A$ is opposite the $67^\circ$ angle.
* So, $A = 67^\circ$.
* Angle B: Angles on a straight line add up to $180^\circ$. Angle $B$ and the $67^\circ$ angle are neighbors on a straight line.
* Calculation: $180^\circ - 67^\circ = 113^\circ$. So, $B = 113^\circ$.
* Angle C: Angle $C$ is vertical to Angle $B$, so they are equal. Alternatively, it is supplementary to Angle $A$.
* So, $C = 113^\circ$.

Problem 3

Goal: Find angles $A$, $B$, and $C$.
* Angle A: Angles on a straight line add to $180^\circ$. Angle $A$ and the $80^\circ$ angle form a straight line.
* Calculation: $180^\circ - 80^\circ = 100^\circ$. So, $A = 100^\circ$.
* Angle B: Angle $B$ and the $80^\circ$ angle are vertical angles (opposite each other).
* So, $B = 80^\circ$.
* Angle C: Angle $C$ and Angle $A$ are vertical angles. Also, Angle $C$ and the $80^\circ$ angle form a straight line.
* Calculation: $180^\circ - 80^\circ = 100^\circ$. So, $C = 100^\circ$.

Problem 4

Goal: Find angles $A$, $B$, and $C$ in the trapezoid.
* Angle A: In an isosceles trapezoid, the base angles are equal. The bottom left angle is $80^\circ$, so the bottom right angle ($A$) is also $80^\circ$.
* So, $A = 80^\circ$.
* Angle B: Consecutive angles between parallel sides (same-side interior angles) add up to $180^\circ$. Angle $B$ and the top-left $100^\circ$ angle are consecutive. Wait, looking at the diagram, the top-left is $100^\circ$ and top-right is $B$. The bottom-left is $80^\circ$. Since the legs are non-parallel, we look at the parallel bases. The angle adjacent to the $80^\circ$ angle on the same leg is $B$? No, usually $A$ and $B$ refer to specific vertices. Let's assume standard labeling: Bottom-Right is $A$, Top-Right is $B$, Top-Left is given as $100^\circ$? No, the label $100^\circ$ is inside the top left corner. The label $80^\circ$ is inside the bottom left corner.
* Let's re-evaluate based on parallel lines. The top and bottom sides are parallel. Therefore, consecutive interior angles add to $180^\circ$.
* Angle $B$ (top right) and Angle $A$ (bottom right) are consecutive interior angles with the left side? No.
* Let's use the property that consecutive angles along a leg sum to $180^\circ$.
* Left leg: Top angle ($100^\circ$) + Bottom angle ($80^\circ$) = $180^\circ$. This confirms the lines are parallel.
* Right leg: Angle $B$ (top) + Angle $A$ (bottom) = $180^\circ$.
* In an isosceles trapezoid, base angles are equal. Bottom Left ($80^\circ$) = Bottom Right ($A$). So $A = 80^\circ$.
* Top Left ($100^\circ$) = Top Right ($B$). So $B = 100^\circ$.
* Angle C: This is the exterior angle at the bottom right vertex. It forms a linear pair with Angle $A$.
* Calculation: $180^\circ - 80^\circ = 100^\circ$. So, $C = 100^\circ$.

Problem 5

Goal: Find angles $A, B, C, D, E, F$.
* Angle A: Vertical to the $50^\circ$ angle.
* So, $A = 50^\circ$.
* Angle B: Forms a straight line with the $50^\circ$ angle.
* Calculation: $180^\circ - 50^\circ = 130^\circ$. So, $B = 130^\circ$.
* Angle C: Vertical to Angle $B$.
* So, $C = 130^\circ$.
* Angle D: Vertical to the $60^\circ$ angle.
* So, $D = 60^\circ$.
* Angle E: Forms a straight line with the $60^\circ$ angle.
* Calculation: $180^\circ - 60^\circ = 120^\circ$. So, $E = 120^\circ$.
* Angle F: Vertical to Angle $E$.
* So, $F = 120^\circ$.

Problem 6

Goal: Find angles $A, B, C, D, E$.
* Angle A: Corresponds to the $120^\circ$ angle? No, let's look at the intersection. The angle vertically opposite to the interior angle on the top line is needed.
* Let's find the interior angle on the top line first. The angle labeled $120^\circ$ and the interior angle next to it form a straight line. Interior angle = $180^\circ - 120^\circ = 60^\circ$.
* Actually, simpler approach: Angle $A$ and the $120^\circ$ angle are consecutive interior angles? No.
* Let's identify relationships directly.
* Angle A: It is in the position of "alternate exterior" or can be found via vertical angles. The angle vertical to the $120^\circ$ angle is also $120^\circ$. That vertical angle and Angle $A$ are consecutive interior angles? No.
* Let's use Corresponding Angles. The angle corresponding to Angle $A$ is the one below the transversal on the top parallel line. The angle supplementary to $120^\circ$ is $60^\circ$. So the interior angle is $60^\circ$. Angle $A$ corresponds to the angle *outside*?
* Let's restart Problem 6 carefully.
* Top intersection: We have an obtuse angle of $120^\circ$. The acute angle next to it on the straight line is $180^\circ - 120^\circ = 60^\circ$.
* Angle A: Angle $A$ is vertically opposite to the $60^\circ$ angle we just found? No, Angle $A$ is outside. Angle $A$ and the $120^\circ$ angle are alternate exterior angles? No.
* Let's look at Angle A's position. It is an acute angle. The angle vertically opposite to the $120^\circ$ angle is $120^\circ$. Angle $A$ and that $120^\circ$ angle form a linear pair. So $A = 180^\circ - 120^\circ = 60^\circ$.
* Angle B: Angle $B$ is vertically opposite to Angle $A$. So $B = 60^\circ$. (Or, Angle $B$ corresponds to the $60^\circ$ interior angle on the top line).
* Angle C: Look at the bottom intersection. We have a triangle formed by the transversal and another line. But wait, there are two parallel horizontal lines cut by a transversal. Then there is another line cutting through.
* Let's look at the labels again. $A, B$ are at the top intersection. $C, D, E$ are at the bottom intersection involving a triangle.
* Okay, Parallel Lines Rule: The interior angle on the top right is $60^\circ$ (supplement of 120). The alternate interior angle at the bottom left (inside the parallel lines) is also $60^\circ$.
* Let's look at the triangle at the bottom. One angle is given as $30^\circ$. Another angle is part of the parallel line intersection.
* The angle vertically opposite to the alternate interior angle ($60^\circ$) is inside the triangle? No.
* Let's trace the transversal (the slanted line going from top-left to bottom-right).
* Top intersection: Exterior angle is $120^\circ$. Interior angle on the same side is $60^\circ$.
* Bottom intersection: The alternate interior angle to the top $60^\circ$ is the angle at the bottom-left of the transversal/parallel-line intersection. So that angle is $60^\circ$.
* Angle C: Angle $C$ is vertically opposite to this $60^\circ$ angle? No, Angle $C$ is adjacent to the triangle.
* Let's look at Angle D. Angle $D$ is inside the triangle. The vertex of the triangle lies on the bottom parallel line.
* The angle formed by the transversal and the bottom parallel line (interior) is $60^\circ$ (alternate interior to the top $60^\circ$).
* This $60^\circ$ angle is composed of Angle $D$ and the $30^\circ$ angle? Or is Angle $D$ adjacent?
* Looking at the diagram: The transversal creates an angle with the parallel line. Inside the triangle, there is a $30^\circ$ angle. The third angle of the triangle is on the parallel line.
* Let's assume the standard "Z" pattern. The alternate interior angle from the top ($60^\circ$) is the entire angle at the bottom vertex between the transversal and the parallel line.
* In the diagram, Angle $D$ and the $30^\circ$ angle seem to make up that alternate interior angle? No, the $30^\circ$ is inside a triangle formed by a *third* line.
* Let's look at Angle E. Angle $E$ is an exterior angle to the triangle?
* Let's try a different path.
* Top Line: Angle supplement to $120^\circ$ is $60^\circ$. This is the interior angle on the right side of the transversal.
* Bottom Line: The consecutive interior angle adds to $180^\circ$. So the interior angle on the right side at the bottom is $180^\circ - 60^\circ = 120^\circ$.
* This $120^\circ$ angle is split into Angle $D$ and... wait.
* Let's look at Angle C. Angle $C$ is vertically opposite to the interior angle on the left?
* Let's simplify.
* Angle A: Corresponds to the angle supplementary to 120? No. Angle $A$ and the $120^\circ$ angle are on a straight line? No, they are around a point. The angle vertically opposite to $120^\circ$ is $120^\circ$. Angle $A$ is adjacent to it on a straight line (the transversal). So $A = 180^\circ - 120^\circ = 60^\circ$.
* Angle B: Vertically opposite to $A$. So $B = 60^\circ$.
* Now move to the bottom intersection. The transversal crosses the bottom parallel line.
* The angle corresponding to Angle $B$ ($60^\circ$) is the angle in the same position at the bottom intersection (top-left of the intersection). Let's call this angle $X$. So $X = 60^\circ$.
* Angle C: Angle $C$ is vertically opposite to Angle $X$? No, Angle $C$ is labeled as the angle vertically opposite to the interior angle on the bottom right?
* Let's look at the triangle. The vertices are:
1. Intersection of transversal and bottom parallel line.
2. A point on the bottom parallel line to the right.
3. An intersection with a third line.
* Actually, it looks like Angle $D$ and the $30^\circ$ angle are adjacent angles that together form the alternate interior angle with the top $60^\circ$ angle?
* If the top interior angle (right side) is $60^\circ$, then the alternate interior angle (bottom left side) is $60^\circ$.
* In the diagram, the angle labeled $D$ and the angle labeled $30^\circ$ are NOT adjacent in a way that sums to the alternate interior.
* Let's look at Angle E. Angle $E$ is an exterior angle of the triangle?
* Let's assume the line forming the $30^\circ$ angle intersects the parallel line.
* Key Insight: The angle vertically opposite to the top $60^\circ$ (interior) is the bottom $60^\circ$ (exterior)? No.
* Let's use Corresponding Angles.
* Top interior right angle = $60^\circ$ (since $180-120=60$).
* Bottom interior right angle (same side) = $120^\circ$ (since $180-60=120$).
* This bottom interior right angle is composed of Angle $D$ and the angle inside the triangle adjacent to it? No.
* Let's look at the "Z". Top left interior angle is $120^\circ$? No, top left exterior is 120, so top left interior is 60? No.
* Top Left Exterior = $120^\circ$. Top Left Interior = $180-120=60^\circ$? No, they are vertical? No, linear pair.
* If Top Left Exterior is 120, then Top Left Interior is $180-120=60$? No, that depends on which side the 120 is. The arc is obtuse. So the interior angle next to it is acute.
* Let's assume the angle marked $120^\circ$ is the Top-Left Exterior angle.
* Then Top-Left Interior = $180 - 120 = 60^\circ$? No, vertical angles. The angle vertically opposite to the exterior 120 is the interior bottom-right of that intersection.
* Let's stick to the simplest interpretation:
* Angle adjacent to $120^\circ$ on the straight line is $60^\circ$. This is the Top-Right Interior angle.
* Therefore, the Bottom-Left Interior angle (alternate interior) is $60^\circ$.
* Looking at the bottom intersection: The angle between the transversal and the parallel line (on the left side, inside) is $60^\circ$.
* Angle $C$ is vertically opposite to this $60^\circ$ angle? No, Angle $C$ is shown as the angle vertically opposite to the *other* side.
* Let's look at Angle D. Angle $D$ is inside a triangle. One vertex is on the parallel line. The angle of the triangle at this vertex is adjacent to the $60^\circ$ alternate interior angle?
* Actually, usually in these problems, the line creating the $30^\circ$ angle is just a random transversal.
* Let's look at Angle E. Angle $E$ is the exterior angle to the triangle at the bottom right vertex?
* Let's try this:
* Top Right Interior Angle = $60^\circ$.
* Bottom Left Interior Angle = $60^\circ$ (Alternate Interior).
* Angle $C$ is vertically opposite to the Bottom Left Interior Angle? If so, $C = 60^\circ$.
* Angle $D$ is part of the angle on the straight line?
* Let's look at the triangle containing the $30^\circ$ angle.
* The sum of angles in a triangle is $180^\circ$.
* We need two angles to find the third.
* One angle is $30^\circ$.
* Another angle is at the intersection of the transversal and the bottom parallel line.
* The angle vertically opposite to the Alternate Interior Angle ($60^\circ$) is $60^\circ$. This angle is "inside" the V shape but outside the parallel strip?
* Let's assume Angle $D$ is the angle inside the triangle at the left vertex.
* The angle vertically opposite to the Alternate Interior Angle ($60^\circ$) is Angle $D$? If so, $D = 60^\circ$.
* If $D = 60^\circ$ and the other angle is $30^\circ$, then the third angle (let's call it $Y$) is $180 - 60 - 30 = 90^\circ$.
* Angle $E$ is supplementary to Angle $Y$? Angle $E$ is shown as the exterior angle at that vertex. So $E = 180 - 90 = 90^\circ$.
* Let's check Angle $C$ again. Angle $C$ is vertically opposite to the angle adjacent to $D$ on the straight line?
* If $D=60$, the angle next to it on the straight line is $120$. Angle $C$ is vertically opposite to that? No.
* Let's restart Problem 6 with the most standard configuration interpretation.
* Angle A: Linear pair with $120^\circ$? No, $A$ and $120$ are not on a line. The line is the transversal. The horizontal line is the parallel line. The angle $120^\circ$ is bounded by the transversal and the parallel line. Angle $A$ is bounded by the transversal and the parallel line. They are vertical angles? No. They are adjacent on the parallel line. So $A + 120 = 180 \rightarrow A = 60^\circ$.
* Angle B: Vertical to $A$. So $B = 60^\circ$.
* Angle C: Corresponding angle to $A$? No. Alternate Exterior to the bottom right?
* Let's find the angle at the bottom intersection corresponding to $B$. $B$ is Top-Left (relative to intersection center). The corresponding angle at the bottom is Top-Left. Let's call it $B'$. $B' = 60^\circ$.
* Angle $C$ is vertically opposite to $B'$? No, Angle $C$ is labeled in the position of the Bottom-Left angle?
* Let's look at the labels $C, D, E$.
* $C$ is the angle vertically opposite to the interior angle on the left?
* Let's assume the angle marked $C$ is the Alternate Interior Angle to the top-right interior angle ($60^\circ$). If so, $C = 60^\circ$.
* However, $C$ is drawn outside the parallel lines.
* Let's look at Angle D. $D$ is inside the triangle. The vertex is on the parallel line. The angle $D$ and the angle marked $30^\circ$ are adjacent?
* Let's assume the line segment forming the right side of the triangle is parallel to the transversal? No indication.
* Let's assume the standard "Triangle Exterior Angle Theorem" or similar.
* Let's go with this likely intended logic:
1. Top intersection: Angle adjacent to $120^\circ$ is $60^\circ$. This is the interior angle on the right.
2. Bottom intersection: The alternate interior angle (on the left) is $60^\circ$.
3. Angle $C$ is vertically opposite to this $60^\circ$ angle? No, Angle $C$ is adjacent to it on the straight line?
4. Let's look at Angle $D$. Angle $D$ is vertically opposite to the alternate interior angle ($60^\circ$). So $D = 60^\circ$.
5. Now consider the triangle with angles: $D$ ($60^\circ$), $30^\circ$, and a third angle $Z$.
6. $Z = 180^\circ - 60^\circ - 30^\circ = 90^\circ$.
7. Angle $E$ is supplementary to $Z$ (forms a linear pair on the bottom parallel line). So $E = 180^\circ - 90^\circ = 90^\circ$.
8. Angle $C$: Angle $C$ is vertically opposite to the angle adjacent to $D$ on the transversal? No. Angle $C$ is vertically opposite to the angle formed by the transversal and the parallel line (bottom left, interior). That interior angle is $60^\circ$ (alternate interior to top right 60? No, consecutive interior sum to 180. Top Right Interior = 60. Bottom Right Interior = 120. Bottom Left Interior = 60).
9. So Bottom Left Interior = $60^\circ$. Angle $C$ is vertically opposite to the Bottom Left Exterior?
10. Actually, looking at the position, Angle $C$ and the Bottom Left Interior Angle ($60^\circ$) form a linear pair on the transversal? No.
11. Angle $C$ is vertically opposite to Angle $D$? No.
12. Let's look at Angle $C$ again. It is in the position of the Corresponding Angle to the top-left angle ($120^\circ$)? If so, $C = 120^\circ$.
13. Let's check if $C=120$ makes sense. If $C=120$, and it's vertical to the bottom-left exterior, then the bottom-left interior is $60$. This matches our calculation.
14. So, $C = 120^\circ$.

Final Answer:
Problem 1:
A = 45°
B = 135°

Problem 2:
A = 67°
B = 113°
C = 113°

Problem 3:
A = 100°
B = 80°
C = 100°

Problem 4:
A = 80°
B = 100°
C = 100°

Problem 5:
A = 50°
B = 130°
C = 130°
D = 60°
E = 120°
F = 120°

Problem 6:
A = 60°
B = 60°
C = 120°
D = 60°
E = 90°