Triangles and Their Angles

Name the parts of a triangle, name triangles by their sides and by their angles, see why the angles of a triangle add up to 180°, and use the exterior angle of a triangle to find unknown angles.

Before you start

An acute angle is less than $90^{\circ}$ , a right angle is exactly $90^{\circ}$ , and an obtuse angle is greater than $90^{\circ}$ and less than $180^{\circ}$ (Lesson 11.1).
Adjacent angles on a straight line add up to $180^{\circ}$ (adj. ∠s on a str. line) (Lesson 11.2).
If $AB \parallel CD$ , alternate angles are equal (alt. ∠s, AB // CD) and corresponding angles are equal (corr. ∠s, AB // CD) (Lesson 11.4).
An equation such as $6x = 180$ is solved by dividing both sides by 6: $x = 30$ (Lesson 9.1).

Sides, Vertices and Angles

A triangle is a closed figure with three straight sides. The three points where the sides meet are its vertices (one vertex, two or more vertices).

A triangle is named by its three vertices: the triangle with vertices $A$ , $B$ and $C$ is written $\triangle ABC$ . Its sides are $AB$ , $BC$ and $CA$ , and its three angles, inside the triangle, are $\angle BAC$ at $A$ , $\angle ABC$ at $B$ and $\angle ACB$ at $C$ . The vertices may be read in any order: $\triangle ABC$ , $\triangle BCA$ and $\triangle CAB$ are the same triangle.

The side opposite an angle is the side that does not touch the angle's vertex. In the figure, the side opposite $\angle BAC$ is $BC$ . In the same way, the angle opposite side $AB$ is $\angle ACB$ , the angle at the vertex that $AB$ does not touch.

Example 1: Sides and angles opposite each other

For $\triangle PQR$ in the figure:

(a) name the side opposite $\angle PQR$ ;

(b) name the angle opposite side $PQ$ ;

Solution

(a) The vertex of $\angle PQR$ is $Q$ (the middle letter). The side that does not touch $Q$ is $PR$ .

(b) Side $PQ$ touches the vertices $P$ and $Q$ . The vertex it does not touch is $R$ , and the angle there is $\angle PRQ$ .

(a) $PR$ (b) $\angle PRQ$ (c) $\angle PQR$ and $\angle PRQ$

Naming Triangles by Their Sides and by Their Angles

Sides of equal length are marked with the same number of small ticks. Sides with no ticks are not marked as equal.

Triangles named by their sides

A scalene triangle has no two sides equal.

An isosceles triangle has at least two sides equal.

An equilateral triangle has all three sides equal.

An equilateral triangle has at least two equal sides, so by the definition it is also isosceles: it is a special isosceles triangle. When a question asks you to name a triangle, give the most specific name, the one that says the most. A triangle with three equal sides is named equilateral.

Triangles named by their angles

An acute-angled triangle has all three angles acute.

A right-angled triangle has one right angle.

An obtuse-angled triangle has one obtuse angle.

To name a triangle by its angles, look at its largest angle: if the largest angle is acute, all three are acute.

Every triangle has a name of each kind. Name it by its sides and by its angles together: "an isosceles right-angled triangle", "a scalene obtuse-angled triangle".

Example 2: Naming triangles

Name each triangle by its sides and by its angles.

(a) $\triangle DEF$ :

(b) $\triangle KLM$ , with $KL = 3$ cm, $LM = 5$ cm, $KM = 7$ cm and $\angle KLM = 120^{\circ}$ :

(c) $\triangle XYZ$ , with $XY = 7$ cm, $YZ = 5$ cm and $XZ = 6$ cm. Its largest angle, $\angle XZY$ , is about $78^{\circ}$ .

Solution

(a) Sides: the ticks show $DE = DF$ , so at least two sides are equal: isosceles. Angles: the square at $D$ marks a right angle: right-angled.

(b) Sides: 3 cm, 5 cm and 7 cm are all different: scalene. Angles: $120^{\circ}$ is greater than $90^{\circ}$ and less than $180^{\circ}$ , so it is obtuse: obtuse-angled.

(c) Sides: 7 cm, 5 cm and 6 cm are all different: scalene. Angles: the largest angle, about $78^{\circ}$ , is acute, so all three angles are acute: acute-angled.

(a) an isosceles right-angled triangle (b) a scalene obtuse-angled triangle (c) a scalene acute-angled triangle

The Angle Sum of a Triangle

In primary school you found that the three angles of a triangle add up to $180^{\circ}$ , for example by tearing the three corners off a paper triangle and fitting them together along a straight line. Tearing or measuring shows it for the triangles you try. Reasoning with the rules of Chapter 11 shows it for every triangle.

Take any $\triangle ABC$ , with its angles $a$ at $A$ , $b$ at $B$ and $c$ at $C$ . Through $A$ , draw a construction line $XY$ parallel to $BC$ .

Why the angles of a triangle add up to 180°

$\angle XAB = \angle ABC = b$ (alt. ∠s, XY // BC)

$\angle YAC = \angle ACB = c$ (alt. ∠s, XY // BC)

At $A$ , the angles $\angle XAB$ , $\angle BAC$ and $\angle YAC$ lie together on the straight line $XY$ , so $b + a + c = 180^{\circ}$ (adj. ∠s on a str. line).

So $a + b + c = 180^{\circ}$ . Nothing here depends on the shape of the triangle, so it is true for every triangle.

The angle sum of a triangle

The sum of the angles of a triangle is $180^{\circ}$ .

In $\triangle ABC$ : $\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$ .

The reason is written (∠ sum of △), short for "the angle sum of a triangle is 180°".

Two right angles already add up to $180^{\circ}$ , leaving nothing for the third angle, and two obtuse angles add up to more than $180^{\circ}$ . So a triangle cannot have two right angles, or two obtuse angles, or one of each. Every triangle has at least two acute angles.

In a right-angled triangle, the other two angles add up to $180^{\circ} - 90^{\circ} = 90^{\circ}$ .

Example 3: The third angle

In $\triangle ABC$ , $\angle ABC = 62^{\circ}$ and $\angle ACB = 51^{\circ}$ . Find $x$ , and name the triangle by its angles.

Solution

$x + 62 + 51 = 180$ (∠ sum of △)

$x = 180 - 62 - 51 = 67$

The three angles are $67^{\circ}$ , $62^{\circ}$ and $51^{\circ}$ . The largest, $67^{\circ}$ , is acute, so all three are acute.

Check: $67^{\circ} + 62^{\circ} + 51^{\circ} = 180^{\circ}$ .

$x = 67$ , so $\angle BAC = 67^{\circ}$ ; $\triangle ABC$ is acute-angled.

Example 4: Naming a triangle from two of its angles

Two angles of a triangle are $23^{\circ}$ and $49^{\circ}$ . Find the third angle and name the triangle by its angles.

Solution

Third angle $= 180^{\circ} - 23^{\circ} - 49^{\circ} = 108^{\circ}$ (∠ sum of △)

$108^{\circ}$ is greater than $90^{\circ}$ and less than $180^{\circ}$ , so it is obtuse.

Check: $23^{\circ} + 49^{\circ} + 108^{\circ} = 180^{\circ}$ .

The third angle is $108^{\circ}$ , and the triangle is obtuse-angled.

Example 5: Can these be the angles of a triangle?

(a) Can a triangle have angles $50^{\circ}$ , $60^{\circ}$ and $80^{\circ}$ ? Explain.

(b) Can a triangle have two of its angles $95^{\circ}$ and $90^{\circ}$ ? Explain.

Solution

(a) $50^{\circ} + 60^{\circ} + 80^{\circ} = 190^{\circ}$ . The angles of a triangle add up to exactly $180^{\circ}$ (∠ sum of △), and $190^{\circ}$ is not $180^{\circ}$ .

(b) $95^{\circ} + 90^{\circ} = 185^{\circ}$ , already more than $180^{\circ}$ before the third angle is added. A triangle has at most one angle that is not acute.

(a) No: the three angles add up to $190^{\circ}$ , not $180^{\circ}$ . (b) No: those two angles alone add up to $185^{\circ}$ , more than $180^{\circ}$ .

Angles Given in Terms of $x$

When the angles of a triangle are given as expressions in $x$ , the angle sum gives an equation: the three expressions add up to 180. Solve it, put the value of $x$ back into each expression, and check that the angles add up to $180^{\circ}$ .

When the angles are in a ratio such as $2 : 3 : 4$ , call them $2x^{\circ}$ , $3x^{\circ}$ and $4x^{\circ}$ and do the same.

Example 6: Angles $x^{\circ}$ , $2x^{\circ}$ and $3x^{\circ}$

The angles of $\triangle PQR$ are $x^{\circ}$ , $2x^{\circ}$ and $3x^{\circ}$ , as shown. Find $x$ and each angle, and name the triangle by its angles.

Solution

$x + 2x + 3x = 180$ (∠ sum of △)

Collect like terms: $6x = 180$

Divide both sides by 6: $x = 30$

$\angle QPR = 30^{\circ}$ , $\angle PQR = 2 \times 30^{\circ} = 60^{\circ}$ and $\angle PRQ = 3 \times 30^{\circ} = 90^{\circ}$ .

Check: $30^{\circ} + 60^{\circ} + 90^{\circ} = 180^{\circ}$ .

One angle is $90^{\circ}$ , a right angle.

$x = 30$ ; the angles are $30^{\circ}$ , $60^{\circ}$ and $90^{\circ}$ , and $\triangle PQR$ is right-angled.

Example 7: Angles in a ratio

The angles of a triangle are in the ratio $2 : 3 : 4$ . Find each angle and name the triangle by its angles.

Solution

Let the angles be $2x^{\circ}$ , $3x^{\circ}$ and $4x^{\circ}$ .

$2x + 3x + 4x = 180$ (∠ sum of △)

Collect like terms: $9x = 180$

Divide both sides by 9: $x = 20$

The angles are $2 \times 20^{\circ} = 40^{\circ}$ , $3 \times 20^{\circ} = 60^{\circ}$ and $4 \times 20^{\circ} = 80^{\circ}$ .

Check: $40^{\circ} + 60^{\circ} + 80^{\circ} = 180^{\circ}$ , and $40 : 60 : 80 = 2 : 3 : 4$ .

The largest angle, $80^{\circ}$ , is acute.

The angles are $40^{\circ}$ , $60^{\circ}$ and $80^{\circ}$ ; the triangle is acute-angled.

The Exterior Angle of a Triangle

The three angles inside a triangle are its interior angles. Carry one side straight on past a vertex, and a new angle appears outside the triangle.

Exterior angle of a triangle

An exterior angle of a triangle is the angle between one side and the extension of an adjacent side.

In the figure, $BC$ is extended to $D$ , and $\angle ACD$ is an exterior angle at $C$ .

An exterior angle and the interior angle beside it lie together on a straight line, so they add up to $180^{\circ}$ (adj. ∠s on a str. line): $\angle ACD + \angle ACB = 180^{\circ}$ .

The interior angles at the other two vertices, $\angle BAC$ and $\angle ABC$ , are the interior opposite angles of $\angle ACD$ .

An exterior angle is less than $180^{\circ}$ . The large reflex angle all the way round the outside of a vertex is not an exterior angle. At each vertex there are two exterior angles, one for each side that can be extended; they are vertically opposite, so they are equal.

Why an exterior angle equals the two interior opposite angles

In the figure, $e + c = 180^{\circ}$ (adj. ∠s on a str. line), so $e = 180^{\circ} - c$ .

Also $a + b + c = 180^{\circ}$ (∠ sum of △), so $a + b = 180^{\circ} - c$ .

Both $e$ and $a + b$ are $180^{\circ}$ take away the same angle $c$ , so $e = a + b$ .

The exterior angle of a triangle

An exterior angle of a triangle is equal to the sum of the two interior opposite angles.

In the figure: $\angle ACD = \angle BAC + \angle ABC$ .

The reason is written (ext. ∠ of △), short for "the exterior angle of a triangle equals the sum of the interior opposite angles".

Example 8: Finding an exterior angle

In $\triangle ABC$ , $BC$ is extended to $D$ . $\angle BAC = 52^{\circ}$ and $\angle ABC = 63^{\circ}$ . Find $x$ .

Solution

The angle marked $x^{\circ}$ is the exterior angle at $C$ ; its interior opposite angles are $\angle BAC$ and $\angle ABC$ .

$x = 52 + 63 = 115$ (ext. ∠ of △)

Check: the interior angle $\angle ACB = 180^{\circ} - 115^{\circ} = 65^{\circ}$ (adj. ∠s on a str. line), and $52^{\circ} + 63^{\circ} + 65^{\circ} = 180^{\circ}$ (∠ sum of △).

$x = 115$ , so $\angle ACD = 115^{\circ}$ .

Example 9: Finding an interior opposite angle

In $\triangle ABC$ , $BC$ is extended to $D$ . $\angle ABC = 47^{\circ}$ and $\angle ACD = 128^{\circ}$ . Find $x$ .

Solution

The exterior angle $\angle ACD$ equals the sum of its interior opposite angles, $\angle BAC$ and $\angle ABC$ :

$x + 47 = 128$ (ext. ∠ of △)

Subtract 47 from both sides: $x = 81$

Check: $\angle ACB = 180^{\circ} - 128^{\circ} = 52^{\circ}$ (adj. ∠s on a str. line), and $81^{\circ} + 47^{\circ} + 52^{\circ} = 180^{\circ}$ (∠ sum of △).

$x = 81$ , so $\angle BAC = 81^{\circ}$ .

Example 10: An exterior angle with an equation

In $\triangle ABC$ , $BC$ is extended to $D$ . $\angle BAC = (2x + 10)^{\circ}$ , $\angle ABC = 3x^{\circ}$ and $\angle ACD = 140^{\circ}$ . Find $x$ and $\angle ACB$ .

Solution

$(2x + 10) + 3x = 140$ (ext. ∠ of △)

Collect like terms: $5x + 10 = 140$

Subtract 10 from both sides: $5x = 130$

Divide both sides by 5: $x = 26$

So $\angle BAC = (2 \times 26 + 10)^{\circ} = 62^{\circ}$ and $\angle ABC = 3 \times 26^{\circ} = 78^{\circ}$ .

$\angle ACB = 180^{\circ} - 140^{\circ} = 40^{\circ}$ (adj. ∠s on a str. line)

Check: $62^{\circ} + 78^{\circ} = 140^{\circ}$ , and $62^{\circ} + 78^{\circ} + 40^{\circ} = 180^{\circ}$ (∠ sum of △).

$x = 26$ and $\angle ACB = 40^{\circ}$ .

Chains of Reasons

In a larger figure, the angle you want may not be found in one step. Look for an angle you can find, with any rule from this lesson or Chapter 11, write it with its reason, and use it to find the next. Say which triangle you are using each time.

Example 11: A triangle and parallel lines

In $\triangle ABC$ , $D$ is on $AB$ and $E$ is on $AC$ , and $DE \parallel BC$ . $\angle ADE = 70^{\circ}$ and $\angle ACB = 45^{\circ}$ . Find $x$ .

Solution

The angle $x^{\circ}$ , $\angle BAC$ , is an angle of $\triangle ABC$ , which has only one other angle marked. Find $\angle ABC$ first: it and $\angle ADE$ are corresponding angles, with $AB$ as the transversal.

$\angle ABC = \angle ADE = 70^{\circ}$ (corr. ∠s, DE // BC)

In $\triangle ABC$ : $x + 70 + 45 = 180$ (∠ sum of △)

$x = 180 - 70 - 45 = 65$

Check: $65^{\circ} + 70^{\circ} + 45^{\circ} = 180^{\circ}$ .

$x = 65$ , so $\angle BAC = 65^{\circ}$ .

Example 12: A roof truss

The frame of a roof is a triangle $ABC$ with a horizontal beam $BC$ . A vertical post $AD$ joins the top $A$ to $D$ on the beam, so $\angle ADC = 90^{\circ}$ . $\angle ABC = 32^{\circ}$ and $\angle DAC = 58^{\circ}$ . Find $x$ and $y$ , and the angle $\angle BAC$ at the top of the roof.

Solution

The post is upright on the horizontal beam, so $\angle ADB = 90^{\circ}$ as well (adj. ∠s on a str. line: $180^{\circ} - 90^{\circ}$ ).

In $\triangle ABD$ : $x + 32 + 90 = 180$ (∠ sum of △), so $x = 180 - 32 - 90 = 58$ .

In $\triangle ADC$ : $y + 58 + 90 = 180$ (∠ sum of △), so $y = 180 - 58 - 90 = 32$ .

$\angle BAC = \angle BAD + \angle DAC = 58^{\circ} + 58^{\circ} = 116^{\circ}$

Check in $\triangle ABC$ : $116^{\circ} + 32^{\circ} + 32^{\circ} = 180^{\circ}$ (∠ sum of △).

$x = 58$ and $y = 32$ ; the angle at the top of the roof is $\angle BAC = 116^{\circ}$ .

Common Mistakes

Naming a triangle with three equal sides "isosceles" when the question asks for its name. It is isosceles, but the most specific name is equilateral.
Thinking a triangle has only one name. Name it by its sides and by its angles: a right-angled triangle can be isosceles, and an obtuse-angled triangle can be scalene.
Deciding that an angle is a right angle because it looks like one. Use the square mark or the angles given, not the look of the drawing.
Using the interior angle beside an exterior angle in the rule. An exterior angle equals the sum of the two interior angles at the other two vertices; the one beside it makes $180^{\circ}$ with it.
Taking the reflex angle outside a vertex as the exterior angle. The exterior angle is between one side and the extension of an adjacent side.
With angles in the ratio $2 : 3 : 4$ , dividing $180^{\circ}$ by 3 (the number of angles) instead of by 9 (the number of parts).
Stopping at $x$ . When the question asks for an angle, put $x$ back into its expression: in Example 10, $x = 26$ , but $\angle BAC = (2x + 10)^{\circ} = 62^{\circ}$ .

Practice

Try each question on paper first, giving a reason for every step, then open its solution.

Practice 1: Sides and angles opposite each other

For $\triangle LMN$ in the figure:

(a) name the side opposite $\angle LMN$ ;

(b) name the angle opposite side $MN$ ;

ShowHide solution

(a) The vertex of $\angle LMN$ is $M$ . The side that does not touch $M$ is $LN$ .

(b) Side $MN$ does not touch $L$ , so the angle opposite it is the angle at $L$ , $\angle MLN$ .

(a) $LN$ (b) $\angle MLN$ (c) $\angle MNL$

Practice 2: Naming triangles

Name each triangle by its sides and by its angles. For (c), name it by its sides only.

(a) $\triangle STU$ :

(b) $\triangle EFG$ , with $EF = 6$ cm, $FG = 8$ cm, $EG = 10$ cm and a right angle at $F$ :

ShowHide solution

(a) The ticks show $ST = SU$ : isosceles. $110^{\circ}$ is obtuse: obtuse-angled.

(b) 6 cm, 8 cm and 10 cm are all different: scalene. The angle at $F$ is a right angle: right-angled.

(a) an isosceles obtuse-angled triangle (b) a scalene right-angled triangle (c) an equilateral triangle

Practice 3: The third angle

Two angles of a triangle are given. Find the third angle and name the triangle by its angles.

(a) $37^{\circ}$ and $53^{\circ}$ (b) $28^{\circ}$ and $47^{\circ}$ (c) $64^{\circ}$ and $58^{\circ}$

ShowHide solution

(a) $180^{\circ} - 37^{\circ} - 53^{\circ} = 90^{\circ}$ (∠ sum of △). One angle is a right angle: right-angled.

(b) $180^{\circ} - 28^{\circ} - 47^{\circ} = 105^{\circ}$ (∠ sum of △). $105^{\circ}$ is obtuse: obtuse-angled.

(c) $180^{\circ} - 64^{\circ} - 58^{\circ} = 58^{\circ}$ (∠ sum of △). The angles are $64^{\circ}$ , $58^{\circ}$ and $58^{\circ}$ ; the largest, $64^{\circ}$ , is acute: acute-angled.

(a) $90^{\circ}$ , right-angled (b) $105^{\circ}$ , obtuse-angled (c) $58^{\circ}$ , acute-angled

Practice 4: Possible or not?

(a) Can a triangle have angles $35^{\circ}$ , $65^{\circ}$ and $70^{\circ}$ ? Explain.

(b) Can a triangle have angles $42^{\circ}$ , $59^{\circ}$ and $79^{\circ}$ ? If it can, name it by its angles.

(c) Siti says, "A triangle can have two obtuse angles if they are only just obtuse, like $91^{\circ}$ and $92^{\circ}$ ." Explain why she is wrong.

ShowHide solution

(a) $35^{\circ} + 65^{\circ} + 70^{\circ} = 170^{\circ}$ , which is not $180^{\circ}$ (∠ sum of △).

(b) $42^{\circ} + 59^{\circ} + 79^{\circ} = 180^{\circ}$ , so these can be the angles of a triangle. The largest, $79^{\circ}$ , is acute: acute-angled.

(c) $91^{\circ} + 92^{\circ} = 183^{\circ}$ , already more than $180^{\circ}$ with the third angle still to come. Any two obtuse angles add up to more than $180^{\circ}$ , so no triangle has two.

(a) No: the angles add up to $170^{\circ}$ . (b) Yes; it is acute-angled. (c) Two obtuse angles add up to more than $180^{\circ}$ ( $91^{\circ} + 92^{\circ} = 183^{\circ}$ ), but all three angles of a triangle add up to exactly $180^{\circ}$ .

Practice 5: A ramp

A ramp $AC$ leads up to a loading platform. The platform's upright side $AB$ is at right angles to the level ground $DCB$ , and the ramp makes an angle of $16^{\circ}$ with the ground. Find $x$ and $y$ .

ShowHide solution

In $\triangle ABC$ : $x + 16 + 90 = 180$ (∠ sum of △), so $x = 180 - 16 - 90 = 74$ .

$\angle ACD$ and $\angle ACB$ lie together on the straight line $DB$ : $y = 180 - 16 = 164$ (adj. ∠s on a str. line).

Check: $\angle ACD$ is an exterior angle of $\triangle ABC$ , so it equals $\angle BAC + \angle ABC = 74^{\circ} + 90^{\circ} = 164^{\circ}$ (ext. ∠ of △).

$x = 74$ and $y = 164$ : $\angle BAC = 74^{\circ}$ and $\angle ACD = 164^{\circ}$ .

Practice 6: An exterior angle

In $\triangle PQR$ , $QR$ is extended to $S$ . $\angle PQR = 58^{\circ}$ and $\angle QPR = 71^{\circ}$ . Find $x$ and $y$ .

ShowHide solution

$\angle PRS$ is the exterior angle at $R$ , and its interior opposite angles are $\angle PQR$ and $\angle QPR$ :

$x = 58 + 71 = 129$ (ext. ∠ of △)

$y = 180 - 129 = 51$ (adj. ∠s on a str. line)

Check: $58^{\circ} + 71^{\circ} + 51^{\circ} = 180^{\circ}$ (∠ sum of △).

$x = 129$ and $y = 51$ : $\angle PRS = 129^{\circ}$ and $\angle PRQ = 51^{\circ}$ .

Practice 7: Three expressions

The angles of $\triangle ABC$ are $x^{\circ}$ , $(x + 20)^{\circ}$ and $(2x + 40)^{\circ}$ , as shown. Find $x$ and each angle, and name the triangle by its angles.

ShowHide solution

$x + (x + 20) + (2x + 40) = 180$ (∠ sum of △)

Collect like terms: $4x + 60 = 180$

Subtract 60 from both sides: $4x = 120$

Divide both sides by 4: $x = 30$

$\angle ABC = 30^{\circ}$ , $\angle ACB = (30 + 20)^{\circ} = 50^{\circ}$ and $\angle BAC = (2 \times 30 + 40)^{\circ} = 100^{\circ}$ .

Check: $30^{\circ} + 50^{\circ} + 100^{\circ} = 180^{\circ}$ . $100^{\circ}$ is obtuse.

$x = 30$ ; the angles are $30^{\circ}$ , $50^{\circ}$ and $100^{\circ}$ , and $\triangle ABC$ is obtuse-angled.

Practice 8: Angles in a ratio

The angles of a triangle are in the ratio $2 : 3 : 7$ . Wei Ming thinks the triangle is acute-angled. Find each angle and say whether he is right.

ShowHide solution

Let the angles be $2x^{\circ}$ , $3x^{\circ}$ and $7x^{\circ}$ .

$2x + 3x + 7x = 180$ (∠ sum of △)

Collect like terms: $12x = 180$

Divide both sides by 12: $x = 15$

The angles are $2 \times 15^{\circ} = 30^{\circ}$ , $3 \times 15^{\circ} = 45^{\circ}$ and $7 \times 15^{\circ} = 105^{\circ}$ .

Check: $30^{\circ} + 45^{\circ} + 105^{\circ} = 180^{\circ}$ .

The largest angle, $105^{\circ}$ , is obtuse.

The angles are $30^{\circ}$ , $45^{\circ}$ and $105^{\circ}$ . Wei Ming is wrong: the triangle is obtuse-angled.

Practice 9: An exterior angle in $x$

In $\triangle ABC$ , $BC$ is extended to $D$ . $\angle BAC = 2x^{\circ}$ , $\angle ABC = (x + 40)^{\circ}$ and $\angle ACD = (5x - 20)^{\circ}$ . Find $x$ and $\angle ACB$ .

ShowHide solution

The exterior angle $\angle ACD$ equals the sum of the interior opposite angles $\angle BAC$ and $\angle ABC$ :

$5x - 20 = 2x + (x + 40)$ (ext. ∠ of △)

Collect like terms: $5x - 20 = 3x + 40$

Subtract $3x$ from both sides: $2x - 20 = 40$

Add 20 to both sides: $2x = 60$

Divide both sides by 2: $x = 30$

So $\angle BAC = 60^{\circ}$ , $\angle ABC = 70^{\circ}$ and $\angle ACD = (5 \times 30 - 20)^{\circ} = 130^{\circ}$ .

$\angle ACB = 180^{\circ} - 130^{\circ} = 50^{\circ}$ (adj. ∠s on a str. line)

Check: $60^{\circ} + 70^{\circ} = 130^{\circ}$ , and $60^{\circ} + 70^{\circ} + 50^{\circ} = 180^{\circ}$ (∠ sum of △).

$x = 30$ and $\angle ACB = 50^{\circ}$ .

Practice 10: The exterior angle with a parallel line

In $\triangle ABC$ , $BC$ is extended to $D$ , and $CE$ is drawn parallel to $BA$ . $\angle BAC = 50^{\circ}$ and $\angle ABC = 72^{\circ}$ .

(a) Find $x$ and $y$ , giving a reason for each.

(b) Hence show that $\angle ACD = \angle BAC + \angle ABC$ .

ShowHide solution

(a) With $AC$ as the transversal, $\angle ACE$ and $\angle BAC$ are alternate angles:

$x = 50$ (alt. ∠s, BA // CE)

With $BD$ as the transversal, $\angle ECD$ and $\angle ABC$ are corresponding angles:

$y = 72$ (corr. ∠s, BA // CE)

(b) The reasons in (a) do not depend on the sizes of the angles: in every triangle drawn like this, $\angle ACE = \angle BAC$ (alt. ∠s, BA // CE) and $\angle ECD = \angle ABC$ (corr. ∠s, BA // CE). So $\angle ACD = \angle ACE + \angle ECD = \angle BAC + \angle ABC$ . Here, $50^{\circ} + 72^{\circ} = 122^{\circ}$ .

This is a second way to see the exterior angle rule: the parallel line $CE$ splits the exterior angle into two parts, one equal to each interior opposite angle.

Check: $\angle ACB = 180^{\circ} - 122^{\circ} = 58^{\circ}$ , and $50^{\circ} + 72^{\circ} + 58^{\circ} = 180^{\circ}$ (∠ sum of △).

(a) $x = 50$ and $y = 72$ . (b) $\angle ACD = \angle ACE + \angle ECD = \angle BAC + \angle ABC$ ; here $50^{\circ} + 72^{\circ} = 122^{\circ}$ .

Practice 11: A set square on a ruler

Mei Ling's set square is a right-angled triangle $ABC$ with $\angle ABC = 90^{\circ}$ and $\angle BAC = 30^{\circ}$ . She stands it on the straight edge of a ruler with $BC$ along the edge, and marks a point $D$ on the edge beyond $C$ . Find $x$ and $y$ .

ShowHide solution

In $\triangle ABC$ : $x + 90 + 30 = 180$ (∠ sum of △), so $x = 180 - 90 - 30 = 60$ .

$y = 180 - 60 = 120$ (adj. ∠s on a str. line)

Check: $\angle ACD$ is an exterior angle of $\triangle ABC$ , so $y = 90 + 30 = 120$ (ext. ∠ of △).

$x = 60$ and $y = 120$ : $\angle ACB = 60^{\circ}$ and $\angle ACD = 120^{\circ}$ .

Practice 12: The side of a footbridge

The side of a footbridge is made of steel triangles between two straight beams, $AE$ at the top and $BD$ at the bottom, with $AE \parallel BD$ . $C$ is on $BD$ . $\angle ABC = 60^{\circ}$ , $\angle ACB = 55^{\circ}$ and $\angle ECD = 65^{\circ}$ . Find $x$ , $y$ and $z$ .

ShowHide solution

In $\triangle ABC$ : $x + 60 + 55 = 180$ (∠ sum of △), so $x = 180 - 60 - 55 = 65$ .

With $AC$ as the transversal, $\angle CAE$ and $\angle ACB$ are alternate angles:

$y = 55$ (alt. ∠s, AE // BD)

At $C$ , the angles $\angle ACB$ , $\angle ACE$ and $\angle ECD$ lie together on the straight line $BD$ :

$\angle ACE = 180^{\circ} - 55^{\circ} - 65^{\circ} = 60^{\circ}$ (adj. ∠s on a str. line)

In $\triangle ACE$ : $z + 55 + 60 = 180$ (∠ sum of △), so $z = 180 - 55 - 60 = 65$ .

Check: $\angle AEC$ and $\angle ECD$ are alternate angles with $CE$ as the transversal, so $z = 65$ (alt. ∠s, AE // BD), as found.

$x = 65$ , $y = 55$ and $z = 65$ : $\angle BAC = 65^{\circ}$ , $\angle CAE = 55^{\circ}$ and $\angle AEC = 65^{\circ}$ .

Key Points

$\triangle ABC$ has vertices $A$ , $B$ , $C$ , sides $AB$ , $BC$ , $CA$ and angles $\angle BAC$ , $\angle ABC$ , $\angle ACB$ ; the side opposite an angle does not touch its vertex.
By sides: scalene (no two sides equal), isosceles (at least two sides equal), equilateral (all three sides equal).
By angles: acute-angled (all three angles acute), right-angled (one right angle), obtuse-angled (one obtuse angle).
Name a triangle by its sides and by its angles, and give the most specific name: three equal sides make it equilateral.
The angles of a triangle add up to $180^{\circ}$ (∠ sum of △). So a triangle has at most one right or obtuse angle.
An exterior angle is between one side and the extension of an adjacent side; with the interior angle beside it, it makes $180^{\circ}$ .
An exterior angle of a triangle equals the sum of the two interior opposite angles (ext. ∠ of △).
With angles in $x$ or in a ratio, form an equation, solve it, find every angle and check the sum.

Before you start

Sides, Vertices and Angles

Example 1: Sides and angles opposite each other

Naming Triangles by Their Sides and by Their Angles

Triangles named by their sides

Triangles named by their angles

Example 2: Naming triangles

The Angle Sum of a Triangle

Why the angles of a triangle add up to 180°

The angle sum of a triangle

Example 3: The third angle

Example 4: Naming a triangle from two of its angles

Example 5: Can these be the angles of a triangle?

Angles Given in Terms of xxx

Example 6: Angles x∘x^{\circ}x∘, 2x∘2x^{\circ}2x∘ and 3x∘3x^{\circ}3x∘

Example 7: Angles in a ratio

The Exterior Angle of a Triangle

Exterior angle of a triangle

Why an exterior angle equals the two interior opposite angles

The exterior angle of a triangle

Example 8: Finding an exterior angle

Example 9: Finding an interior opposite angle

Example 10: An exterior angle with an equation

Chains of Reasons

Example 11: A triangle and parallel lines

Example 12: A roof truss

Common Mistakes

Practice

Practice 1: Sides and angles opposite each other

Practice 2: Naming triangles

Practice 3: The third angle

Practice 4: Possible or not?

Practice 5: A ramp

Practice 6: An exterior angle

Practice 7: Three expressions

Practice 8: Angles in a ratio

Practice 9: An exterior angle in xxx

Practice 10: The exterior angle with a parallel line

Practice 11: A set square on a ruler

Practice 12: The side of a footbridge

Key Points

Angles Given in Terms of $x$

Example 6: Angles $x^{\circ}$ , $2x^{\circ}$ and $3x^{\circ}$

Practice 9: An exterior angle in $x$