IGCSE Maths: Alternate Angles and Triangle Angle Sum (Parallel Lines)

Two parallel lines $PQ$ and $RS$ are drawn, with $PQ$ above $RS$ .

A straight line (a transversal) crosses the upper line $PQ$ at the point $A$ and crosses the lower line $RS$ at the point $B$ . A third point $C$ lies on the lower line $RS$ , to the right of $B$ , so that $A$ , $B$ and $C$ form triangle $ABC$ .

At $A$ , the angle measured from $PQ$ (towards $Q$ , the left-hand end) to the line $AB$ is angle $QAB = 58^\circ$ .

At $C$ , the angle of the triangle, angle $ACB$ , is $47^\circ$ .

(The two parallel lines are $PQ$ and $RS$ ; $AB$ is the transversal cutting them, forming a "Z" shape with $A$ on $PQ$ and $B$ on $RS$ .)

(a) Write down the size of angle $ABC$ , the angle between the line $RS$ and the line $BA$ . Give a geometrical reason for your answer. [2]

(b) Work out the size of angle $BAC$ . Give a geometrical reason for your answer. [2]

Understanding the set-up

We are told that $PQ \parallel RS$ and that the line through $A$ and $B$ is a transversal crossing both parallel lines. The points $A$ , $B$ and $C$ form a triangle, with $B$ and $C$ both on the lower line $RS$ .

The information given is:

angle $QAB = 58^\circ$ , at $A$ , between the parallel line $PQ$ and the transversal $AB$ ;
angle $ACB = 47^\circ$ , the angle of the triangle at $C$ .

Part (a): the size of angle $ABC$

Look at the transversal $AB$ cutting the two parallel lines. It makes a “Z” shape:

one arm of the Z is along $PQ$ at $A$ (towards $Q$ ), giving angle $QAB = 58^\circ$ ;
the other arm of the Z is along $RS$ at $B$ (towards $C$ ), giving angle $ABC$ .

These two marked angles sit inside the parallel lines and on opposite sides of the transversal. Angles in this position are called alternate angles, and alternate angles between parallel lines are equal.

$\text{angle } ABC = \text{angle } QAB = 58^\circ$

Reason: alternate angles (between parallel lines $PQ$ and $RS$ ) are equal.

$\boxed{\text{angle } ABC = 58^\circ}$

Part (b): the size of angle $BAC$

Now work inside triangle $ABC$ . Its three interior angles are:

angle $ABC = 58^\circ$ (found in part (a)),
angle $ACB = 47^\circ$ (given),
angle $BAC = {?}$ (the one we want).

The angles in any triangle add up to $180^\circ$ :

$\text{angle } BAC + \text{angle } ABC + \text{angle } ACB = 180^\circ$

$\text{angle } BAC + 58^\circ + 47^\circ = 180^\circ$

$\text{angle } BAC + 105^\circ = 180^\circ$

$\text{angle } BAC = 180^\circ - 105^\circ = 75^\circ$

Reason: the angles in a triangle add up to $180^\circ$ .

$\boxed{\text{angle } BAC = 75^\circ}$

Quick check

Adding the three triangle angles:

$75^\circ + 58^\circ + 47^\circ = 180^\circ \checkmark$

The total is $180^\circ$ , which confirms the answers are consistent.

A note on naming the parallel-line reason

In part (a) the equal angles form a Z (alternate angles). Be careful not to call them corresponding angles (which form an F) or co-interior angles (a C, which would instead add to $180^\circ$ ). Choosing the wrong name (or working out $180^\circ - 58^\circ = 122^\circ$ for a co-interior pair) is the most common error here.

Worked solution

Understanding the set-up

Part (a): the size of angle ABCABCABC

Part (b): the size of angle BACBACBAC

Quick check

A note on naming the parallel-line reason

Part (a): the size of angle $ABC$

Part (b): the size of angle $BAC$