Calculating Angles in Triangles Worksheet | 4th Grade PDF Worksheets - Free Printable

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

Section A: Calculate the missing angles

*Rule: The angles inside any triangle always add up to $180^\circ$.*

1)
* Given angles: $100^\circ$ and $30^\circ$.
* Add them: $100 + 30 = 130$.
* Subtract from 180: $180 - 130 = 50$.
* $a = 50$

2)
* Given angles: $50^\circ$ and $45^\circ$.
* Add them: $50 + 45 = 95$.
* Subtract from 180: $180 - 95 = 85$.
* $b = 85$

3)
* This is a right-angled triangle (the square symbol means $90^\circ$).
* Given angles: $90^\circ$ and $60^\circ$.
* Add them: $90 + 60 = 150$.
* Subtract from 180: $180 - 150 = 30$.
* $c = 30$

4)
* The tick marks show this is an isosceles triangle, so the two bottom angles ($d$) are equal.
* Top angle: $34^\circ$.
* Subtract top angle from 180: $180 - 34 = 146$.
* Divide by 2 to find the base angles: $146 \div 2 = 73$.
* $d = 73$

5)
* The tick marks show this is an isosceles triangle. The sides with ticks are equal, which means the angles opposite them are equal.
* The angle opposite the left ticked side is $e$. The angle opposite the right ticked side is the one at the bottom right. Wait, looking closely at diagram 5, the tick marks are on the two legs meeting at the top vertex? No, the tick marks are on the bottom side and the right side. Let's look closer.
* Actually, usually in these diagrams, if two sides have a single tick mark, the angles *opposite* those sides are equal.
* Side with tick 1 is the bottom side. Angle opposite is the top angle.
* Side with tick 2 is the right side. Angle opposite is the bottom-left angle ($26^\circ$).
* Therefore, the top angle is also $26^\circ$.
* So we have $26^\circ$ and $26^\circ$.
* Sum: $26 + 26 = 52$.
* Remaining angle $e$: $180 - 52 = 128$.
* $e = 128$

6)
* All three sides have tick marks. This is an equilateral triangle.
* All angles in an equilateral triangle are equal.
* $180 \div 3 = 60$.
* $f = 60$

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Section B: Calculate the missing angles

*Rules: Angles on a straight line add to $180^\circ$. Vertically opposite angles are equal. Exterior angle of a triangle equals sum of two interior opposite angles.*

1)
* Find the interior angle next to $116^\circ$ first. They are on a straight line.
* $180 - 116 = 64^\circ$.
* Now use the triangle sum rule for angles $41^\circ$, $64^\circ$, and $a$.
* $41 + 64 = 105$.
* $180 - 105 = 75$.
* $a = 75$
*(Alternative method: Exterior angle $116 = 41 + a$. So $a = 116 - 41 = 75$.)*

2)
* First, find the third angle inside the triangle. It's a right triangle ($90^\circ$) with another angle of $49^\circ$.
* $180 - 90 - 49 = 41^\circ$.
* Angle $b$ and this $41^\circ$ angle are on a straight line.
* $180 - 41 = 139$.
* $b = 139$
*(Alternative method: Exterior angle $b$ equals sum of interior opposites: $90 + 49 = 139$.)*

3)
* The angle vertically opposite to $93^\circ$ is inside the triangle. So the top internal angle is $93^\circ$.
* Triangle angles: $93^\circ$, $40^\circ$, and $c$.
* $93 + 40 = 133$.
* $180 - 133 = 47$.
* $c = 47$

4)
* Look at the triangle on the right side first. We know two angles: $10^\circ$ and $38^\circ$.
* Let's find the third angle (let's call it $y$) which is on the straight line base.
* $180 - 10 - 38 = 132^\circ$.
* Now look at the angle adjacent to $y$ on the straight line. Let's call it $z$.
* $180 - 132 = 48^\circ$. This is the bottom-right angle of the left triangle.
* Now solve for $d$ in the left triangle. Angles are $62^\circ$, $48^\circ$, and $d$.
* $62 + 48 = 110$.
* $180 - 110 = 70$.
* $d = 70$

5)
* This is an isosceles triangle (tick marks on two sides).
* The angle between the two equal sides is $56^\circ$.
* The other two angles are equal. Let's call them $y$.
* $180 - 56 = 124$.
* $124 \div 2 = 62$. So the base angles are $62^\circ$.
* Angle $e$ is supplementary to one of these base angles (they form a straight line along the side of the larger triangle structure? No, looking at diagram 5, $e$ is an exterior angle to the small triangle formed by the bisector? Or is it simply part of a larger triangle?)
* Let's re-examine diagram 5 carefully. It shows a large triangle split into two. The left part has a vertical side and a horizontal base? No.
* Let's look at the markings again. There is a large triangle. A line is drawn from the top vertex to the base.
* The left small triangle has a top angle part unknown, bottom left angle unknown, bottom right angle unknown.
* Wait, the tick marks are on the *left side* of the big triangle and the *internal dividing line*. This means the triangle on the left is isosceles.
* The top angle of that left triangle is NOT given directly. The angle $56^\circ$ is the whole top angle? Or just the left part? The arc for $56^\circ$ seems to cover only the left part of the top vertex.
* Let's assume the angle labeled $56^\circ$ is the top angle of the left-hand triangle.
* Since the left side and the middle line have ticks, the left triangle is isosceles with the vertex at the bottom left? No, the ticks are on the side lengths.
* Side 1: Left edge of the big triangle.
* Side 2: The internal line.
* These two sides meet at the top vertex.
* Therefore, the angles opposite these sides are equal.
* Angle opposite Left Edge is the bottom-right angle of the left triangle.
* Angle opposite Internal Line is the bottom-left angle of the left triangle.
* So, Bottom-Left Angle = Bottom-Right Angle (of the left sub-triangle).
* Let these be $y$.
* Top angle is $56^\circ$.
* $2y + 56 = 180 \rightarrow 2y = 124 \rightarrow y = 62^\circ$.
* So the bottom-left angle of the whole shape is $62^\circ$. The angle inside the left triangle at the base is $62^\circ$.
* Now, what is $e$? $e$ is the angle adjacent to the top angle $56^\circ$? No, $e$ is marked as the angle inside the right sub-triangle at the top vertex.
* Wait, the label $e^\circ$ is inside the right-hand triangle, at the top vertex.
* Is there more information? The right-hand triangle has a tick mark on its right side. And the base looks like a straight line.
* Usually, if there are no other markings, we might assume the big triangle is isosceles too? Or maybe the right triangle is also isosceles?
* Let's look at the tick marks again.
* Left side of big triangle: 1 tick.
* Middle line: 1 tick.
* Right side of big triangle: 1 tick.
* Ah! All three relevant segments have single tick marks.
* This means: Left Side = Middle Line = Right Side.
* This creates TWO isosceles triangles.
* Left Triangle: Sides are Equal. Top angle is $56^\circ$. As calculated before, the base angles are $(180-56)/2 = 62^\circ$.
* So, the angle at the bottom center (where the middle line meets the base) for the left triangle is $62^\circ$.
* Consequently, the angle at the bottom center for the right triangle is $180 - 62 = 118^\circ$ (angles on a straight line).
* Right Triangle: The middle line and the right side are equal (both have 1 tick). This makes the right triangle isosceles too.
* The angles opposite these equal sides must be equal.
* Angle opposite Middle Line is the bottom-right corner of the big triangle.
* Angle opposite Right Side is the top angle of the right triangle, which is $e$.
* So, $e$ = Bottom-Right Angle.
* The sum of angles in the right triangle is $180^\circ$. The angles are:
1. Bottom-Center: $118^\circ$.
2. Top ($e$).
3. Bottom-Right (also $e$).
* Equation: $118 + e + e = 180$.
* $2e = 180 - 118$.
* $2e = 62$.
* $e = 31$.
* $e = 31$

6)
* This involves two triangles touching at a vertex (vertically opposite angles).
* Top Triangle:
* We have an exterior angle of $147^\circ$. The interior angle next to it is $180 - 147 = 33^\circ$.
* The other given angle is $53^\circ$.
* The third angle (at the intersection point) is $180 - 53 - 33 = 94^\circ$.
* Bottom Triangle:
* The angle at the intersection point is vertically opposite to the one we just found, so it is also $94^\circ$.
* We have an exterior angle of $138^\circ$. The interior angle next to it is $180 - 138 = 42^\circ$.
* The remaining angle is $f$.
* $f + 94 + 42 = 180$.
* $f + 136 = 180$.
* $f = 44$.
* $f = 44$

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Extension: Regular Pentagon

* A regular pentagon has 5 equal sides and 5 equal interior angles.
* Formula for interior angle of a regular polygon: $(n-2) \times 180 / n$.
* For a pentagon ($n=5$): $(5-2) \times 180 / 5 = 3 \times 180 / 5 = 540 / 5 = 108^\circ$.
* So, every interior angle of the pentagon is $108^\circ$.
* The diagram shows a triangle formed by drawing a diagonal.
* The angle labeled $72^\circ$ is outside the pentagon. It is supplementary to the interior angle?
* $180 - 108 = 72^\circ$. Yes, the line extends straight out from the bottom side.
* We need to find angle $x$.
* Let's identify the triangle containing $x$. It is formed by two diagonals or a diagonal and a side?
* Looking at the pink shape, it's a triangle inside the pentagon.
* The vertices of this triangle are:
1. The bottom-right vertex of the pentagon.
2. The top vertex of the pentagon.
3. The bottom-left vertex of the pentagon.
* Wait, let's trace the lines.
* One line goes from the bottom-left vertex to the top vertex.
* Another line goes from the bottom-right vertex to the top vertex.
* This forms an isosceles triangle with the top vertex of the pentagon.
* The angle at the top vertex of the pentagon is $108^\circ$.
* The triangle formed by the two diagonals from the top vertex to the bottom vertices is an isosceles triangle because the diagonals of a regular pentagon are equal in length? Or rather, the sides of the pentagon are equal, so the triangle formed by connecting three consecutive vertices (like Top, Bottom-Right, Bottom-Left?? No, that's not consecutive).
* Let's look at the specific triangle shaded pink.
* Its vertices appear to be:
1. The top vertex of the pentagon.
2. The bottom-right vertex.
3. The bottom-left vertex.
* If this is the case, the sides of this triangle are two diagonals and one side (the bottom base)? No, the bottom base is a side of the pentagon. The other two sides are diagonals.
* In a regular pentagon, the triangle formed by connecting three vertices (skipping none? i.e., adjacent vertices) would use two sides and one diagonal.
* The diagram shows the triangle using the bottom side and two diagonals connecting to the top vertex? No, that would span the whole height.
* Let's look at angle $x$. It is located at the bottom-right vertex.
* The angle $x$ is part of the full interior angle ($108^\circ$).
* The triangle shown is formed by the diagonal from bottom-left to top, and the diagonal from bottom-right to top?
* Actually, simpler view: The triangle consists of the Top Vertex, the Bottom-Right Vertex, and the Bottom-Left Vertex.
* This is an isosceles triangle. The two long sides are diagonals. The short base is the bottom side of the pentagon.
* Wait, the diagonals from the top vertex go to the two bottom vertices.
* Let's calculate the angles of this specific triangle (Top, Bottom-Right, Bottom-Left).
* This interpretation assumes the lines are diagonals.
* However, there is a standard property: The diagonal of a regular pentagon trisects the interior angle? No, it splits it into $36^\circ$, $36^\circ$, $36^\circ$? No.
* Let's calculate from scratch.
* Consider the triangle formed by three consecutive vertices, e.g., Top, Top-Right, Bottom-Right.
* This is an isosceles triangle with two sides being sides of the pentagon (length $s$) and the third being a diagonal.
* The angle between the two sides is the interior angle $108^\circ$.
* The other two angles are equal: $(180 - 108) / 2 = 36^\circ$.
* So, the angle between a side and a diagonal is $36^\circ$.
* Now look at angle $x$ in the diagram.
* Angle $x$ is bounded by the bottom side of the pentagon and a diagonal going to the top vertex.
* Let's verify which diagonal. The line goes from the bottom-right vertex to the top vertex.
* This diagonal skips one vertex (the top-right one).
* So, we are looking at the triangle formed by: Bottom-Right Vertex, Top-Right Vertex, and Top Vertex.
* In this triangle (let's call vertices $C, D, E$ starting from bottom-left clockwise $A,B,C,D,E$):
* We are looking at triangle $BCE$? No.
* Let's label vertices clockwise: 1(Top), 2(Top Right), 3(Bottom Right), 4(Bottom Left), 5(Top Left).
* The triangle shaded is defined by vertices 1, 3, and 4?
* The line goes from 4 to 1. The line goes from 3 to 1. The base is 4 to 3.
* So the triangle is $\triangle 1-3-4$.
* This triangle is isosceles because side 1-3 is a diagonal and side 1-4 is a diagonal. In a regular pentagon, all diagonals are equal length.
* So $\triangle 1-3-4$ is isosceles with legs = diagonal length. Base = side length.
* We need the angle at vertex 3 (which is $x$).
* Let's find the angle $\angle 1-3-4$.
* We know the full interior angle at vertex 3 is $108^\circ$.
* This angle is composed of $\angle 1-3-2$ (angle between diagonal 1-3 and side 3-2) and $\angle 2-3-4$ (angle between side 3-2 and side 3-4... wait, 3-4 is the base).
* Actually, easier way:
* Consider the triangle formed by vertices 1, 2, and 3. This uses two sides of the pentagon (1-2 and 2-3) and one diagonal (1-3).
* Angle at vertex 2 is $108^\circ$.
* Triangle 1-2-3 is isosceles.
* Angles at 1 and 3 are $(180 - 108) / 2 = 36^\circ$.
* So, $\angle 1-3-2 = 36^\circ$.
* The full interior angle at vertex 3 is $\angle 1-3-4 + \angle 1-3-2$? No.
* The vertices are 4-3-2-1.
* The interior angle is $\angle 4-3-2 = 108^\circ$.
* The diagonal is 1-3.
* So $\angle 4-3-2$ is split into $\angle 4-3-1$ (which is $x$) and $\angle 1-3-2$ (which we found is $36^\circ$).
* Therefore, $x + 36 = 108$.
* $x = 108 - 36$.
* $x = 72$.

* Let's double check.
* Is $x$ the angle between the diagonal and the side? Yes.
* In a regular pentagon, the diagonals from one vertex divide the interior angle into three equal parts of $36^\circ$ each?
* Diagonal to adjacent vertex (side): $0^\circ$ offset.
* Diagonal to next vertex: $36^\circ$ from the side.
* Diagonal to next vertex: $36^\circ + 36^\circ = 72^\circ$ from the first side?
* Let's re-evaluate.
* Interior angle $108^\circ$.
* Triangle with 2 sides (isosceles): Base angles $36^\circ$. So angle between diagonal and side is $36^\circ$.
* There are two such diagonals from any vertex.
* They split the $108^\circ$ angle into $36^\circ$, $36^\circ$, and $36^\circ$.
* Angle $x$ is shown between the bottom side and the diagonal going to the top vertex.
* The diagonal going to the "nearest" non-adjacent vertex creates a $36^\circ$ angle with the side.
* The diagonal going to the "farthest" vertex (in a hexagon? No, pentagon has only 2 diagonals per vertex).
* From Bottom-Right (Vertex 3):
* Diagonal to Top-Left (Vertex 5). Angle with side 3-4 is $36^\circ$.
* Diagonal to Top (Vertex 1). Angle with side 3-2 is $36^\circ$.
* The middle angle is $108 - 36 - 36 = 36^\circ$.
* Which diagonal is drawn? The line goes to the Top Vertex (Vertex 1).
* So we are looking for the angle between Side 3-4 (Bottom) and Diagonal 3-1.
* We know Angle(Side 3-2, Diagonal 3-1) = $36^\circ$.
* We know Angle(Side 3-2, Side 3-4) = $108^\circ$.
* Therefore, Angle(Side 3-4, Diagonal 3-1) = $108 - 36 = 72^\circ$.
* So $x = 72$.

Final Answer:
Section A
1) a = 50
2) b = 85
3) c = 30
4) d = 73
5) e = 128
6) f = 60

Section B
1) a = 75
2) b = 139
3) c = 47
4) d = 70
5) e = 31
6) f = 44

Extension
x = 72