Angles and Triangles
Theory
Angles on a straight line add to 180°, angles at a point add to 360°, and the three angles of any triangle add to 180°. This Year 9 maths page revises the Year 7 angle facts that Year 9 problem solving relies on: complementary and supplementary angles, vertically opposite angles, angle bisectors, isosceles triangles, the exterior angle theorem and classifying triangles.
Two angles are complementary if they add to \(90^\circ\), and supplementary if they add to \(180^\circ\).
Angles on a straight line add to \(180^\circ\). Angles at a point (all the way round) add to \(360^\circ\). A reflex angle is bigger than \(180^\circ\) and less than \(360^\circ\); an angle and the reflex angle the other way round the same point add to \(360^\circ\). When two straight lines cross, the angles opposite each other are vertically opposite and are equal.
A ray that bisects an angle cuts it into two equal halves.
The angle sum of a triangle is \(180^\circ\). By its angles, a triangle is acute-angled (all angles less than \(90^\circ\)), right-angled (one angle of \(90^\circ\)) or obtuse-angled (one angle more than \(90^\circ\)). By its sides, it is equilateral (three equal sides), isosceles (two equal sides) or scalene (no equal sides). Equal sides are marked with matching ticks.
In an isosceles triangle, the two angles opposite the equal sides are equal. They are the base angles. The third angle is the apex. In an equilateral triangle every angle is \(60^\circ\). It also works backwards: if two angles of a triangle are equal, the sides opposite them are equal, so the triangle is isosceles.
Extend one side and the angle outside is an exterior angle. It equals the sum of the two interior angles furthest from it.
The angle facts, with the reason to write beside each step:
| Fact | Rule | Reason to write |
|---|---|---|
| Complementary | add to \(90^\circ\) | complementary angles |
| Straight line | add to \(180^\circ\) | angles on a straight line |
| At a point | add to \(360^\circ\) | angles at a point |
| Reflex angle | \(360^\circ\) minus the angle the short way round | angles at a point |
| X crossing | equal | vertically opposite angles |
| Triangle | add to \(180^\circ\) | angle sum of a triangle |
| Isosceles | base angles equal | base angles of an isosceles triangle |
| Exterior angle | = sum of the two interior opposite angles | exterior angle of a triangle |
| By its angles | By its sides | ||
|---|---|---|---|
| acute-angled | all three less than \(90^\circ\) | equilateral | all three sides equal |
| right-angled | one angle of \(90^\circ\) | isosceles | two sides equal |
| obtuse-angled | one angle more than \(90^\circ\) | scalene | no sides equal |
How to find an unknown angle
- Mark every angle you know on the diagram, including equal angles from equal sides and halves from a bisector.
- Look for the shape that links the unknown to known angles: a right angle, a straight line, a point, an X crossing or a triangle.
- Write one step at a time with its reason. If the angles have pronumerals, write an equation.
- Solve, substitute back if the question asks for an angle, and check the total (\(90^\circ\), \(180^\circ\) or \(360^\circ\)).
Triangles
- Two angles known: subtract their sum from \(180^\circ\).
- Isosceles: find the base angles with \(\dfrac{180^\circ - \text{apex}}{2}\), or the apex with \(180^\circ - 2 \times \text{base angle}\).
- An angle outside the triangle: use the exterior angle theorem, or the straight angle next to it.
- To classify, give one name from the angles and one from the sides, such as an obtuse-angled isosceles triangle.
(i) Find the complement and the supplement of \(57^\circ\).
(ii) \(AOB\) is a straight line. Find \(x\) and the size of each angle.
(iii) Three angles at a point are \(140^\circ\), \(95^\circ\) and \(y^\circ\). Find \(y\).
(i) Complementary angles add to \(90^\circ\); supplementary angles add to \(180^\circ\).
| \(\text{complement}\) | \(=\) | \(90^\circ - 57^\circ = \mathbf{33^\circ}\) |
| \(\text{supplement}\) | \(=\) | \(180^\circ - 57^\circ = \mathbf{123^\circ}\) |
(ii) The three angles make the straight line \(AOB\), so they add to \(180^\circ\).
| \(2x + (3x + 5) + 25\) | \(=\) | \(180\) |
| \(5x + 30\) | \(=\) | \(180\) |
| \(5x\) | \(=\) | \(150\) |
| \(x\) | \(=\) | \(\mathbf{30}\) |
The angles are \(2(30) = 60^\circ\), \(3(30) + 5 = 95^\circ\) and \(25^\circ\). Check: \(60 + 95 + 25 = 180\) \(\checkmark\)
(iii) Angles at a point add to \(360^\circ\).
| \(y\) | \(=\) | \(360 - (140 + 95)\) |
| \(\) | \(=\) | \(\mathbf{125}\) |
(i) Find \(a = \angle AOD\), \(b = \angle AOC\) and \(c = \angle BOD\), giving a reason for each.
(ii) \(OP\) bisects \(\angle AOC\). Find \(\angle POC\).
(i) Work round the crossing one angle at a time.
| \(a\) | \(=\) | \(\mathbf{64^\circ}\ \text{(vertically opposite } \angle BOC)\) |
| \(b\) | \(=\) | \(180^\circ - 64^\circ = \mathbf{116^\circ}\) |
| \(\text{(straight line } AOB)\) | ||
| \(c\) | \(=\) | \(\mathbf{116^\circ}\ \text{(vertically opposite } b)\) |
Check: \(64 + 116 + 64 + 116 = 360\) \(\checkmark\) (angles at a point)
(ii) A bisector cuts the angle into two equal halves.
| \(\angle POC\) | \(=\) | \(\dfrac{116^\circ}{2} = \mathbf{58^\circ}\) |
(i) The angles of a triangle are \(x\), \(x + 20^\circ\) and \(2x\). Find \(x\) and each angle.
(ii) An isosceles triangle has an apex angle of \(34^\circ\). Find each base angle.
(iii) Another isosceles triangle has base angles of \(38^\circ\). Find its apex angle.
(i) Add the three angles and set the total equal to \(180^\circ\).
| \(x + (x + 20) + 2x\) | \(=\) | \(180\) |
| \(4x + 20\) | \(=\) | \(180\) |
| \(4x\) | \(=\) | \(160\) |
| \(x\) | \(=\) | \(\mathbf{40}\) |
The angles are \(40^\circ\), \(60^\circ\) and \(80^\circ\). Check: \(40 + 60 + 80 = 180\) \(\checkmark\)
(ii) The base angles are equal. Call each one \(x\).
| \(34 + x + x\) | \(=\) | \(180\) |
| \(2x\) | \(=\) | \(146\) |
| \(x\) | \(=\) | \(\mathbf{73^\circ}\) |
(iii) Take both base angles from \(180^\circ\).
| \(\text{apex}\) | \(=\) | \(180^\circ - 2 \times 38^\circ\) |
| \(\) | \(=\) | \(\mathbf{104^\circ}\) |
(i) \(x\), the size of \(\angle ABC\)
(ii) \(\angle ACB\)
(iii) Classify \(\triangle ABC\) by its angles and its sides.
(i) The exterior angle equals the sum of the two interior opposite angles.
| \(52^\circ + x\) | \(=\) | \(115^\circ\) |
| \(x\) | \(=\) | \(\mathbf{63^\circ}\) |
(ii) \(\angle ACB\) and \(\angle BCD\) form a straight angle, so they add to \(180^\circ\).
| \(\angle ACB\) | \(=\) | \(180^\circ - 115^\circ = \mathbf{65^\circ}\) |
Check: \(52 + 63 + 65 = 180\) \(\checkmark\)
(iii) All three angles are less than \(90^\circ\), and no two are equal, so no two sides are equal. \(\triangle ABC\) is an acute-angled scalene triangle.
Common pitfalls
Frequently asked questions
What is the difference between complementary and supplementary angles?
Complementary angles add to \(90^\circ\). Supplementary angles add to \(180^\circ\). The complement of \(57^\circ\) is \(33^\circ\) and its supplement is \(123^\circ\).
What are vertically opposite angles?
When two straight lines cross, the angles opposite each other at the crossing are vertically opposite. They are always equal.
What do the angles in a triangle add up to?
The three angles inside any triangle add up to \(180^\circ\).
How do you find the base angles of an isosceles triangle?
Take the apex angle away from \(180^\circ\), then halve the result. With an apex of \(34^\circ\), each base angle is \(146 \div 2 = 73^\circ\).
What is the exterior angle theorem?
An exterior angle of a triangle equals the sum of the two interior angles that are not next to it.
Can a triangle be both right-angled and isosceles?
Yes. A right-angled isosceles triangle has angles of \(90^\circ\), \(45^\circ\) and \(45^\circ\), with the two shorter sides equal.