A sum of 360
🎯 In this topic you will
- Use the fact that the sum of the angles around a point is 360°
- Show and use the fact that the angles of any quadrilateral add up to 360°
🧠 Key Words
- quadrilateral
- sum
Show Definitions
- quadrilateral: A polygon with exactly four sides and four angles.
- sum: The total obtained by adding two or more numbers or quantities together.
Angles on a Straight Line and Around a Point
The sum of the angles on a straight line is $180^\circ$.
$45^\circ + 72^\circ + 63^\circ = 180^\circ$
A whole turn is $360^\circ$. The sum of the angles around a point is $360^\circ$.
$65^\circ + 53^\circ + 107^\circ + 135^\circ = 360^\circ$
You can apply your algebra skills to find unknown angles, represented by letters.
The sum of the angles of a triangle is $180^\circ$.
A quadrilateral has four straight sides and four angles.
You can draw a straight line to divide the quadrilateral into two triangles.
The six angles of the two triangles make the angles of the quadrilateral.
The sum of the angles of each triangle is $180^\circ$.
The sum of the angles of the quadrilateral is $2 \times 180^\circ = 360^\circ$.
This result is true for any quadrilateral.
You can use the geometrical properties of shapes to calculate missing angles.
❓ EXERCISES
1. Work out the size of the angle that has a letter.
a.
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b.
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c.
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d.
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2. Calculate the size of each angle that has a letter.
a.
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b.
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c.
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d.
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❓ EXERCISES
3. The angles in each of these diagrams are all the same size. What is the size of each angle?
a.
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b.
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4. Calculate the size of angle $B$ in each of these triangles.
a.
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b.
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c.
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5. Three angles of a quadrilateral are $60^\circ$, $80^\circ$ and $110^\circ$. Work out the fourth angle.
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6. In these quadrilaterals, calculate the size of the angles that have a letter.
a.
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b.
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c.
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7. All the angles of a quadrilateral are equal. What can you say about the quadrilateral?
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8. Sofia measures three of the angles of a quadrilateral. Sofia says the angles are $125^\circ$, $160^\circ$ and $90^\circ$.
a. Show that she has made a mistake.
b. Show your answer to part a to another learner. Is your answer clear? Could you improve your answer?
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9. One angle of a quadrilateral is $160^\circ$. The other angles are all the same size. Work out the size of the other three angles.
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10. This shape is a parallelogram. Work out angles $x$, $y$ and $z$.
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❓ EXERCISES
11. $ABCD$ is a quadrilateral. Angle $A=60^\circ$ and angle $B=50^\circ$. Calculate angles $C$ and $D$.
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The interior angle of the quadrilateral at $C$ is a straight-line supplement to $\angle ACB$, hence $C = 180^\circ - 70^\circ = 110^\circ$.
Angles in a quadrilateral sum to $360^\circ$, so $D = 360^\circ - (A + B + C) = 360^\circ - (60^\circ + 50^\circ + 110^\circ) = 140^\circ$.
🧠 Think like a Mathematician
12. All the angles of a quadrilateral are multiples of $30^\circ$.
a. When all the angles are different, show that there is only one possible set of angles.
b. If one of the angles is $90^\circ$, find the other three angles. Show that you have found all possible answers.
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- a. Let the four distinct angles be multiples of $30^\circ$ less than $180^\circ$: $30^\circ,60^\circ,90^\circ,120^\circ,150^\circ$. We need four different values whose sum is $360^\circ$. The only choice is $30^\circ+60^\circ+120^\circ+150^\circ=360^\circ$. Any other selection either repeats an angle or exceeds $360^\circ$, so the set is unique (up to order).
- b. Suppose one angle is $90^\circ$. Let the other three be $30a^\circ,30b^\circ,30c^\circ$ with $1\le a,b,c\le 5$ (angles < $180^\circ$). Then $30(a+b+c)=360^\circ-90^\circ=270^\circ$, so $a+b+c=9$. All unordered integer solutions with $1\le a,b,c\le 5$ are:
$\{1,3,5\}\Rightarrow\{30^\circ,90^\circ,150^\circ\}$Therefore, with one right angle the complete sets (up to order) are: $\{90^\circ,30^\circ,90^\circ,150^\circ\}$, $\{90^\circ,30^\circ,120^\circ,120^\circ\}$, $\{90^\circ,60^\circ,60^\circ,150^\circ\}$, $\{90^\circ,60^\circ,90^\circ,120^\circ\}$, $\{90^\circ,90^\circ,90^\circ,90^\circ\}$. These exhaust all possibilities because every multiple of $30^\circ$ less than $180^\circ$ has been accounted for via the integer solutions to $a+b+c=9$.
$\{1,4,4\}\Rightarrow\{30^\circ,120^\circ,120^\circ\}$
$\{2,2,5\}\Rightarrow\{60^\circ,60^\circ,150^\circ\}$
$\{2,3,4\}\Rightarrow\{60^\circ,90^\circ,120^\circ\}$
$\{3,3,3\}\Rightarrow\{90^\circ,90^\circ,90^\circ\}$
❓ EXERCISES
13. This is a rectangle. Work out the angles that have a letter.
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The angle between the two slanted lines is fixed everywhere. At the top edge of the rectangle it equals $180^\circ-(30^\circ+40^\circ)=110^\circ$. Hence the angle between those same two slanted lines at the upper interior point is also $110^\circ$. Around that point: $$a+b+140^\circ+110^\circ=360^\circ\ .$$ Using the straight–line pair with the $140^\circ$ wedge gives the complementary $40^\circ$ next to it, so the remaining wedge between the two slanted lines is $110^\circ-40^\circ=70^\circ$. Therefore $$a=70^\circ,\qquad b=150^\circ.$$ At the lower edge, the two corner wedges are $20^\circ$ each, so the angle between the same two slanted lines there is $180^\circ-(20^\circ+20^\circ)=140^\circ$. Around the lower interior point the three angles satisfy $$c+d+e=360^\circ-140^\circ=220^\circ.$$ Using the straight–line complements with the upper point configuration gives the two small wedges of $40^\circ$ each next to the vertical, leaving the remaining angle along the base as $220^\circ-40^\circ-40^\circ=100^\circ$. Hence $$c=40^\circ,\quad d=40^\circ,\quad e=100^\circ.$$
❓ EXERCISES
14. Here are two identical triangles.
You can put the triangles together to make a quadrilateral, as shown.
a. i. Find the angles of this quadrilateral.
a. ii. Show that the sum of the angles is $360^\circ$.
b. Find all the different ways of putting the two triangles together to make a quadrilateral. You can turn the triangle over, as shown, if you prefer.
c. i. Find the angles of your quadrilaterals.
c. ii. Show that the sum is $360^\circ$ for each quadrilateral.
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a. i. Each triangle is a $30^\circ$–$60^\circ$–$90^\circ$ right triangle. In the shown assembly, the two small wedges at opposite corners are each $60^\circ$, one corner is a right angle $90^\circ$, and the remaining corner is the supplement needed to make $360^\circ$: $360^\circ-(60^\circ+60^\circ+90^\circ)=150^\circ$. Angles (going around): $60^\circ,\ 150^\circ,\ 60^\circ,\ 90^\circ$.
a. ii. Check: $60^\circ+150^\circ+60^\circ+90^\circ=360^\circ$.
b & c. i. Possible distinct quadrilaterals from two identical $30^\circ$–$60^\circ$–$90^\circ$ triangles (up to rotation/reflection):
- Join along the hypotenuse (two right triangles back-to-back) → a rectangle. Angles: $90^\circ,\ 90^\circ,\ 90^\circ,\ 90^\circ$.
- Join along the longer leg (the side opposite $30^\circ$) → a parallelogram. Adjacent angles: $60^\circ$ and $120^\circ$ (so the set is $60^\circ,\ 120^\circ,\ 60^\circ,\ 120^\circ$).
- Join along the shorter leg (the side opposite $60^\circ$) → another parallelogram. Adjacent angles: $30^\circ$ and $150^\circ$ (so the set is $30^\circ,\ 150^\circ,\ 30^\circ,\ 150^\circ$).
- The arrangement shown in part a (legs not colinear) → angles $60^\circ,\ 150^\circ,\ 60^\circ,\ 90^\circ$ (from part a.i).
c. ii. In every case, the interior angles of the quadrilateral sum to $360^\circ$: $4\times90^\circ=360^\circ$, $60^\circ+120^\circ+60^\circ+120^\circ=360^\circ$, $30^\circ+150^\circ+30^\circ+150^\circ=360^\circ$, and from part a $60^\circ+150^\circ+60^\circ+90^\circ=360^\circ$.