Angle Theorems Worked ExamplesIGCSE Maths 0580

(a) Angles on a straight line add to $180^{\circ}$.

(b) Angles at a point add to $360^{\circ}$.

Co-interior (allied) angles lie between the parallel lines on the same side of the transversal, and they add to $180^{\circ}$.

Corresponding angles sit in matching positions at each intersection and are equal.

(a) The exterior angle equals the sum of the two interior opposite angles.

(b) Find the third interior angle from the angle sum.

The exterior angle and that interior angle are on a straight line.

The two base angles of an isosceles triangle are equal, so subtract the apex from $180^{\circ}$ and halve.

Halve the remainder.

(a) Angles in a quadrilateral sum to $360^{\circ}$.

(b) A diagonal splits any quadrilateral into two triangles, each with an angle sum of $180^{\circ}$.

(a) The three angles lie on a straight line, so they sum to $180^{\circ}$.

Collect the $x$ terms and the constants separately.

Solve.

(b) Substitute back into each expression.

Check the three angles sum correctly.

(a) No parallel-line rule applies to $x$ directly, because $x$ is not formed by a single transversal. Draw a line through $E$ parallel to $PQ$ (and so also parallel to $RS$). It splits $x$ into an upper part and a lower part.

The upper part and angle $QBE$ are alternate angles between the new line and $PQ$, so they are equal.

In the same way, the lower part and angle $SDE$ are alternate angles between the new line and $RS$.

Add the two parts.

(b) The same construction shows that angle $BED$ is always angle $QBE$ + angle $SDE$ for this shape. Work backwards.

(a) $AD = DC$, so triangle $ADC$ is isosceles and its base angles $DAC$ and $DCA$ are equal. Share the remaining $180^{\circ} - 110^{\circ}$ between them.

(b) $AB$ is parallel to $DC$ and $AC$ crosses both lines. Angle $BAC$ and angle $DCA$ are alternate angles.

(c) Use the angle sum of triangle $ABC$.

(d) At $C$, angle $BCD = 73^{\circ} + 35^{\circ} = 108^{\circ}$. $B$, $C$ and $E$ lie on a straight line.

Check with a different fact. $DC$ is parallel to $AB$ and $BE$ crosses both, so angles $DCE$ and $ABC$ are corresponding angles and should be equal. Both are $72^{\circ}$, so the chain of working is consistent.

(a) Angles $AGE$ and $EGB$ lie on the straight line $AB$, so they sum to $180^{\circ}$.

Work out the three marked angles.

(b) We cannot use a parallel-line fact yet — that would assume what we are trying to show. Instead, find angle $BGH$. It is vertically opposite angle $AGE$.

Angles $BGH$ and $GHD$ are co-interior angles between $AB$ and $CD$. They sum to $180^{\circ}$, so the lines must be parallel.

(c) $GK$ bisects angle $BGH$, so it cuts it into two equal halves.

Now that we know $AB$ is parallel to $CD$, use it. Angles $GKH$ and $KGB$ are alternate angles.

Triangle $GHK$ has two equal angles, at $G$ and at $K$, so it is isosceles, with $HG = HK$. Check: the third angle is $180^{\circ} - 40^{\circ} - 40^{\circ} = 100^{\circ}$, which is angle $GHD$, as it should be.

$p$: The $64^{\circ}$ angle and $p$ are in matching positions at the two parallel lines (both above the line, on the right of the same straight line).

$q$: The $131^{\circ}$ angle and $q$ are both between the parallel lines, on the same side of the second straight line, so they are co-interior angles.

$r$: Use the triangle at the bottom. Its angles are $p$, $q$ and the lower angle at the crossing point.

$r$ is vertically opposite that angle.

First find the interior angle of the quadrilateral at $N$. $KNP$ is a straight line.

Now use the angle sum of the quadrilateral $KLMN$.

$KLQ$ is a straight line, so find angle $MLQ$ in the triangle.

$LM = LQ$, so the equal angles of triangle $LMQ$ are the ones opposite those sides: angle $LQM$ and angle $LMQ$. Share what is left of $180^{\circ}$ between them.

In triangle $PRS$, angle $PRS = 90^{\circ}$. Find angle $PSR$.

$RSQ$ is a straight line, so $x$ and angle $PSR$ sum to $180^{\circ}$.

Now use triangle $PSQ$.

Check with the whole triangle $PQR$: angle $RPQ = 180 - 90 - 35 = 55$, and $28 + 27 = 55$. ✓