8.03a–8.03c - Angles on lines | OCR GCSE Mathematics

A single marked angle can determine several others without a protractor. Use turns, straight lines and parallel lines to calculate missing angles, explaining the geometrical reason for each step.

Angles at a point and on a line

Two parallel rails are crossed by a diagonal bar. If one angle is $64^\circ$ , can you find the other seven angles? We will reach that question after establishing two totals which make angle calculations possible.

Imagine facing in one direction, turning through several adjacent angles, and ending where you started. You have made one complete turn. Stopping in the opposite direction instead gives half a turn.

[DIAGRAM: asset_name: A half turn and a full turn; asset_slug: half-and-full-turns; caption: Add adjacent angles which exactly fill the stated turn.; description: The first panel has a straight line AOB with two rays above it dividing the half turn into angles a, b and c. The second has three rays OP, OQ and OR dividing a complete turn into adjacent angles p, q and r. Arcs identify all regions, no region is counted twice.]
Add adjacent angles which exactly fill the stated turn.

Angle totals

Adjacent angles at a point on one side of a straight line sum to $180^\circ$ . Angles filling one complete turn at a point sum to $360^\circ$ .

a+b+c=180^\circ,\qquad p+q+r=360^\circ

These totals are not on the 2027 formula sheet: learn them.

The angles must fill the turn without gaps or overlaps. Two angles which merely share a vertex need not total $180^\circ$ .

Worked example 1: filling a straight angle

Question. In the half-turn diagram, $a=47^\circ$ and $b=68^\circ$ . Find $c$ .

Plan. Subtract both known angles from the straight-line total.

\begin{aligned} c &= 180^\circ-47^\circ-68^\circ && \text{angles on a straight line} \\ &=65^\circ && \text{subtract the known total} \end{aligned}

Answer. $c=65^\circ$ .

Check. $47+68+65=180$ , so the three angles exactly fill the half-turn.

Change only $a$ . Since the total stays fixed, making one angle ten degrees larger must make the missing angle ten degrees smaller.

Worked example 2: a larger known angle

Question. In the same half-turn arrangement, $a=57^\circ$ and $b=68^\circ$ . Find $c$ .

Plan. Use the same straight-line total with the new value of $a$ .

\begin{aligned} c &=180^\circ-57^\circ-68^\circ && \text{angles on a straight line} \\ &=55^\circ && \text{subtract} \end{aligned}

Answer. $c=55^\circ$ .

Check. $57+68+55=180$ . Compared with the first example, $a$ rose by ten and $c$ fell by ten.

For a full turn, change the total rather than the subtraction method.

Worked example 3: filling a full turn

Question. In the full-turn diagram, $p=138^\circ$ and $q=90^\circ$ . Find $r$ .

Plan. Remove the known angles from a complete turn.

\begin{aligned} r &=360^\circ-138^\circ-90^\circ && \text{angles at a point} \\ &=132^\circ && \text{subtract} \end{aligned}

Answer. $r=132^\circ$ .

Check. $138+90+132=360$ , and the remaining angle is positive and smaller than a straight angle.

Now change only $p$ , keeping the quarter-turn $q$ .

Worked example 4: the same complete turn

Question. In the full-turn arrangement, $p=148^\circ$ and $q=90^\circ$ . Find $r$ .

Plan. Subtract the new known total from $360^\circ$ .

\begin{aligned} r&=360^\circ-148^\circ-90^\circ && \text{angles at a point} \\ &=122^\circ && \text{subtract} \end{aligned}

Answer. $r=122^\circ$ .

Check. $148+90+122=360$ ; the ten-degree increase in $p$ has reduced $r$ by ten degrees.

Choose the total from the geometry before doing arithmetic.

Watch out

For the complete-turn diagram, $r=180^\circ-p-q$ uses the wrong total. A straight line is half a turn; the three labelled regions fill a full turn.

Use the full-turn arrangement for this fresh calculation.

Vertically opposite angles

When two straight lines cross, each passes through the intersection rather than stopping there. Opposite angles then have a special equality.

[DIAGRAM: asset_name: Four angles at an intersection; asset_slug: intersecting-angles; caption: Opposite regions are equal, neighbouring regions on a line total 180 degrees.; description: Straight lines AOB and COD cross at O. Going anticlockwise from ray OB, the four regions are labelled a, b, c and d. Angles a and c are vertically opposite, as are b and d. The diagram is schematic, with labelled rays and no numerical measurements.]
Opposite regions are equal, neighbouring regions on a line total 180 degrees.

Vertically opposite angles

Vertically opposite angles are the opposite angles formed by two intersecting straight lines. They are equal: $a=c$ and $b=d$ . “Vertically” does not mean they must appear above and below each other on the page.

Why are they equal? Both $a+b$ and $b+c$ are straight angles. Thus $a+b=180^\circ=b+c$ . Subtract the common $b$ to obtain $a=c$ . The same argument proves $b=d$ . Adjacent angles are supplementary, meaning they sum to $180^\circ$ ; they are not generally equal.

Worked example 5: two different angle sizes

Question. In the intersection diagram, $a=63^\circ$ . Find $c$ and $b$ .

Plan. Use opposite-angle equality first, then the straight line.

\begin{aligned} c&=63^\circ && \text{vertically opposite to }a \\ b&=180^\circ-63^\circ && \text{adjacent angles on a line} \\ &=117^\circ \end{aligned}

Answer. $c=63^\circ$ and $b=117^\circ$ .

Check. The fourth angle is also $117^\circ$ . All four total $63+117+63+117=360$ degrees.

Change only the value of $a$ ; the opposite pairs stay in the same positions.

Worked example 6: equality comes from position

Question. In the same arrangement, $a=73^\circ$ . Find $c$ and $b$ .

Plan. Copy $a$ to its opposite, then subtract from $180^\circ$ .

\begin{aligned} c&=73^\circ && \text{vertically opposite angles} \\ b&=180^\circ-73^\circ && \text{straight line} \\ &=107^\circ \end{aligned}

Answer. $c=73^\circ$ and $b=107^\circ$ .

Check. $73+107=180$ . The acute pair grew by ten degrees while the obtuse pair shrank by ten.

Do not copy an angle into a neighbouring region just because both touch the same intersection.

Watch out

The line $b=a$ is not justified by vertically opposite angles: $b$ and $a$ are adjacent. Only when the lines are perpendicular are all four angles $90^\circ$ .

State both the value and the relevant equality.

Angles on parallel lines

A transversal is a line crossing two other lines. When those two lines are parallel, their common direction creates further equal angles.

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8.03a–8.03c - Angles on lines

Angles at a point and on a line

Vertically opposite angles

Angles on parallel lines

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