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A sum of 360

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visibility 92update 8 months agobookmarkshare

🎯 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.

📘 Worked example

Here are three angles around a point.

Answer:

$142^\circ + 77^\circ = 219^\circ$

The sum of the three angles is $360^\circ$, so $a = 360^\circ - 219^\circ = 141^\circ$.

Add the two known angles: $142^\circ + 77^\circ = 219^\circ$.

Subtract this total from $360^\circ$ to find $a$: $360^\circ - 219^\circ = 141^\circ$.

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.

📘 Worked example

Three of the angles of a quadrilateral are each equal to $85^\circ$. Work out the fourth angle.

Answer:

$3 \times 85^\circ = 255^\circ$

The sum of three of the angles is $255^\circ$.

All four angles add up to $360^\circ$.

The fourth angle is $360^\circ - 255^\circ = 105^\circ$.

Multiply the three equal angles: $3 \times 85^\circ = 255^\circ$.

Since the sum of all four angles in a quadrilateral is $360^\circ$, subtract the total of the three known angles: $360^\circ - 255^\circ = 105^\circ$.

📘 Worked example

This shape is a kite. Calculate the missing angles.

Answer:

There is a vertical line of symmetry, so $a=80^\circ$.

The four angles add up to $360^\circ$.

$135^\circ + 80^\circ + 80^\circ = 295^\circ$, so $b = 360^\circ - 295^\circ = 65^\circ$.

In a kite, the equal sides create a line of symmetry, making the base angles on each side equal; hence $a=80^\circ$.

Angles in any quadrilateral total $360^\circ$. Add the known angles and subtract from $360^\circ$ to find $b$: $b=360^\circ-(135^\circ+80^\circ+80^\circ)=65^\circ$.

🧠 PROBLEM-SOLVING Strategy

Finding Unknown Angles in Lines, Points & Quadrilaterals

Use these steps to solve the Exercise $5.1$–$14$ style problems efficiently.

Mark the givens. Copy each known angle next to its arc and label target letters ($a$, $b$, $x$, …). Note right-angle squares = $90^\circ$.
Apply single-point/line facts first.
- On a straight line: sum = $180^\circ$.
- Around a point: sum = $360^\circ$.
- Vertically opposite angles are equal.
Use triangle facts. In any triangle, angles sum to $180^\circ$. In right triangles, the two acute angles sum to $90^\circ$.
Use quadrilateral facts. Interior angles sum to $360^\circ$. For special quads you might meet:
- Rectangle: every angle $90^\circ$.
- Parallelogram: opposite angles equal; adjacent angles supplementary ($180^\circ$).
- Kite with symmetry: base angles on the symmetric sides are equal.
Work from constraints near the letters. Start where two arcs and one unknown share a point/line, forming equations like $\text{unknown}=180^\circ-\text{(knowns)}$ or $\text{unknown}=360^\circ-\text{(knowns)}$.
Create mini-equations and solve sequentially. For example, with three knowns around a point: $a=360^\circ-(\alpha+\beta+\gamma)$. Propagate each result to adjacent lines/triangles/quadrilaterals.
Check special corner wedges. If a diagram shows little corner numbers like $30^\circ$, $40^\circ$, or $20^\circ$ at the rectangle edges, first find the angle between the two intersecting lines at that corner: edge angle = $180^\circ - (\text{two corner wedges})$. Reuse that same line–line angle at any intersection of those two lines.
Finish with a global check. Verify each solved figure sums correctly: lines $\to 180^\circ$, points $\to 360^\circ$, triangles $\to 180^\circ$, quadrilaterals $\to 360^\circ$. If any sum is off, revisit the last step.

Template equations you’ll use:

Straight line: $x=180^\circ-(p+q)$ | Around a point: $x=360^\circ-(p+q+r)$ | Triangle: $x=180^\circ-(p+q)$ | Quadrilateral: $x=360^\circ-(p+q+r)$

❓ EXERCISES

1. Work out the size of the angle that has a letter.

👀 Show answer

Angles on a straight line sum to $180^\circ$. $a = 180^\circ - 116^\circ = 64^\circ$.

👀 Show answer

Angles around a point sum to $360^\circ$. $b = 360^\circ - 55^\circ - 180^\circ = 125^\circ$.

👀 Show answer

Angles on a straight line sum to $180^\circ$. $c = 180^\circ - (60^\circ + 24^\circ) = 96^\circ$.

👀 Show answer

The right angle is $90^\circ$. Angles on a straight line sum to $180^\circ$. $d = 180^\circ - (90^\circ + 34^\circ) = 56^\circ$.

2. Calculate the size of each angle that has a letter.

👀 Show answer

Angles around a point sum to $360^\circ$. $a = 360^\circ - (130^\circ + 120^\circ) = 110^\circ$.

👀 Show answer

$b = 360^\circ - (155^\circ + 37^\circ) = 168^\circ$.

👀 Show answer

$c = 360^\circ - (68^\circ + 52^\circ + 36^\circ) = 204^\circ$.

👀 Show answer

The right angle is $90^\circ$. Angles on a straight line sum to $180^\circ$. $d = 180^\circ - (90^\circ + 42^\circ) = 48^\circ$.

❓ EXERCISES

3. The angles in each of these diagrams are all the same size. What is the size of each angle?

👀 Show answer

Angles around a point sum to $360^\circ$. Each angle $= 360^\circ \div 3 = 120^\circ$.

👀 Show answer

Angles around a point sum to $360^\circ$. Each angle $= 360^\circ \div 5 = 72^\circ$.

4. Calculate the size of angle $B$ in each of these triangles.

👀 Show answer

Angles in a triangle sum to $180^\circ$. $B = 180^\circ - (57^\circ + 49^\circ) = 74^\circ$.

👀 Show answer

$B = 180^\circ - 90^\circ - 28^\circ = 62^\circ$.

👀 Show answer

$B = 180^\circ - (25^\circ + 38^\circ) = 117^\circ$.

5. Three angles of a quadrilateral are $60^\circ$, $80^\circ$ and $110^\circ$. Work out the fourth angle.

👀 Show answer

Sum of angles in a quadrilateral = $360^\circ$. Fourth angle = $360^\circ - (60^\circ + 80^\circ + 110^\circ) = 110^\circ$.

6. In these quadrilaterals, calculate the size of the angles that have a letter.

👀 Show answer

$a = 360^\circ - (95^\circ + 63^\circ + 110^\circ) = 92^\circ$.

👀 Show answer

$b = 360^\circ - (40^\circ + 35^\circ + 62^\circ) = 223^\circ$.

👀 Show answer

$c = 360^\circ - (35^\circ + 100^\circ + 172^\circ) = 53^\circ$.

7. All the angles of a quadrilateral are equal. What can you say about the quadrilateral?

👀 Show answer

Each angle = $360^\circ \div 4 = 90^\circ$. The quadrilateral is a rectangle (or a square if all sides are equal).

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?

👀 Show answer

Sum of the given angles = $125^\circ + 160^\circ + 90^\circ = 375^\circ$, which is greater than $360^\circ$. This is impossible for a quadrilateral.

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.

👀 Show answer

Let each of the other three equal angles be $x$. $160^\circ + 3x = 360^\circ$ $3x = 200^\circ$ $x = 66.\overline{6}^\circ$.

10. This shape is a parallelogram. Work out angles $x$, $y$ and $z$.

👀 Show answer

In a parallelogram, opposite angles are equal and adjacent angles are supplementary. $x = 68^\circ$, $y = 112^\circ$, $z = 112^\circ$.

❓ EXERCISES

11. $ABCD$ is a quadrilateral. Angle $A=60^\circ$ and angle $B=50^\circ$. Calculate angles $C$ and $D$.

👀 Show answer

Draw the diagonal $AC$. In $\triangle ABC$ the angles sum to $180^\circ$, so $\angle ACB = 180^\circ - (60^\circ + 50^\circ) = 70^\circ$.
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.

👀 show answer

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:
$\{1,3,5\}\Rightarrow\{30^\circ,90^\circ,150^\circ\}$
$\{1,4,4\}\Rightarrow\{30^\circ,120^\circ,120^\circ\}$
$\{2,2,5\}\Rightarrow\{60^\circ,60^\circ,150^\circ\}$
$\{2,3,4\}\Rightarrow\{60^\circ,90^\circ,120^\circ\}$
$\{3,3,3\}\Rightarrow\{90^\circ,90^\circ,90^\circ\}$
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 $a+b+c=9$.

❓ EXERCISES

13. This is a rectangle. Work out the angles that have a letter.

👀 Show answer

Let the two slanted lines from the top corners meet the vertical line at the upper point (where one angle is $140^\circ$) and at the lower point (where the three angles are $c,d,e$).

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 $$c=40^\circ,\quad d=40^\circ,\quad e=100^\circ.$$

Final answers: $a=70^\circ,\ b=150^\circ,\ c=40^\circ,\ d=40^\circ,\ e=100^\circ$.

❓ 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.

👀 Show answer

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 $60^\circ+150^\circ+60^\circ+90^\circ=360^\circ$.

📘 What we've learned

Angles on a straight line add up to $180^\circ$.
Angles around a point sum to $360^\circ$.
Vertically opposite angles are always equal.
The sum of the angles in a triangle is $180^\circ$, and in a quadrilateral is $360^\circ$.
Special quadrilateral properties: in rectangles each angle is $90^\circ$, and in parallelograms opposite angles are equal while adjacent angles sum to $180^\circ$.
We applied these facts systematically to find unknown angles in complex diagrams involving intersecting lines, triangles, and quadrilaterals.

A sum of 360

🎯 In this topic you will

🧠 Key Words

Angles on a Straight Line and Around a Point

🧠 PROBLEM-SOLVING Strategy

Finding Unknown Angles in Lines, Points & Quadrilaterals

❓ EXERCISES

❓ EXERCISES

❓ EXERCISES

🧠 Think like a Mathematician

❓ EXERCISES

❓ EXERCISES

📘 What we've learned

Related Past Papers

Related Tutorials

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