Angles in triangles · Angles, triangles & polygons

Three facts about triangles answer most questions:

The angles of a triangle add up to $180^\circ$ .
An exterior angle (made by producing one side) equals the sum of the two interior angles opposite it.
In an isosceles triangle (two equal sides), the angles opposite the equal sides, the base angles, are equal. In an equilateral triangle every angle is $60^\circ$ .

Angle suma + b + c = 180°

Exterior angleThe exterior angle is a + b

Isosceles triangleEqual sides (ticks) face equal angles

Try it

The angles of a triangleDrag the corners

59°A62°B59°C180°A + B + C121°exterior angle at C

However you drag it, A + B + C = 180°, and the exterior angle at C equals the two interior angles opposite it: 59° + 62° = 121°.

Drag the corners into any shape: the three angles always add to $180^\circ$ , and the exterior angle at $C$ is always $A + B$ . Press “Show why”: the dashed line through $C$ is parallel to $BA$ , so the exterior angle splits into a copy of $A$ (alternate angles) and a copy of $B$ (corresponding angles), the facts from the last lesson.

Isosceles triangles

The equal angles are the ones opposite the equal sides. When the angle between the equal sides is given, the other two share what’s left of $180^\circ$ equally.

Chains of isosceles triangles

Some diagrams have several equal sides in a row. Take the triangles one at a time: each pair of equal sides gives a pair of equal base angles, and the exterior angle of one triangle becomes a base angle of the next.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q12 (a)

In the diagram, $\angle QMN = 34^\circ$ , $|MN| = |NQ| = |QO|$ , $M$ , $N$ , $O$ , $P$ lie on a straight line and $\angle QOP = x$ . Find the value of $x$ .

The first isosceles triangle
$|MN| = |NQ|$ , so the angles opposite them are equal: $\angle NQM = \angle NMQ = 34^\circ$ .
Think first. MN = NQ. Which two angles of triangle MNQ are equal?
The exterior angle at N
$\angle QNO$ is an exterior angle of triangle $MNQ$ , so it equals the two opposite interior angles: $34^\circ + 34^\circ = 68^\circ$ .
Think first. ∠QNO is outside triangle MNQ. What does it equal?
The second isosceles triangle
$|NQ| = |QO|$ , so $\angle QON = \angle QNO = 68^\circ$ (base angles).
Think first. NQ = QO. Which angle equals ∠QNO?
The angle x
$x = \angle QOP = 180^\circ - 68^\circ = 112^\circ$ (angles on a straight line).
Think first. N, O and P are on a straight line.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q2 (b)

In the diagram, $P$ , $U$ , $T$ , $S$ lie on a straight line and $P$ , $Q$ , $R$ on another. $\angle STQ = m$ , $\angle TUQ = 80^\circ$ , $\angle UPQ = r$ , $\angle PQU = n$ and $\angle RQT = 88^\circ$ . Find the value of $(m + n)$ .

The values of r, m and n are not given; the drawing uses r = 20°.

One exterior angle
$\angle TUQ$ is an exterior angle of triangle $PUQ$ : $r + n = 80^\circ$ . (1)
Think first. ∠TUQ = 80° is outside triangle PUQ. What does it equal?
Another exterior angle
The interior angle at $T$ is $180^\circ - m$ (angles on a straight line). $\angle RQT$ is an exterior angle of triangle $PQT$ : $r + (180^\circ - m) = 88^\circ$ , so $m - r = 92^\circ$ . (2)
Think first. ∠RQT = 88° is outside triangle PQT. Which two angles does it equal? One is at T.
Add the equations
$(r + n) + (m - r) = 80^\circ + 92^\circ$ . The $r$ cancels: $m + n = 172^\circ$ .
Think first. You want m + n, and r is unknown. What happens if you add (1) and (2)?

Your turn

WAEC 2023 · Paper 2 · Q12 (a)

(a)
In the diagram, $\overline{PQ} \parallel \overline{MN}$ , $|PR| = |QR|$ , $\angle QMN = 53^\circ$ and $\angle MNP = \left(\frac32 y - 13\right)^\circ$ . Find: (i) $\angle PRQ$ ; (ii) the value of $y$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

$PQ \parallel MN$ , and $QM$ crosses both, so the alternate angles are equal: $\angle PQR = \angle QMN = 53^\circ$ .
$|PR| = |QR|$ , so triangle $PQR$ is isosceles and $\angle QPR = \angle PQR = 53^\circ$ .
Then $\angle PRQ = 180^\circ - 53^\circ - 53^\circ$
$= 74^\circ$ .

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