C4.6; E4.6 - Angles at points and in triangles

C4.6; E4.6 - Angles at points and in triangles

An angle can be forced by the lines around it without being measured. You will calculate angles at points, at crossing lines, and inside triangles and quadrilaterals, giving a geometric reason for each step.

Angles at a point and on a line

Two lines cross and one angle is 68∘68^\circ. Why can you find all four angles without a protractor? Begin with the amount of turning available around their common vertex.

Three-letter notation identifies an angle precisely. In ∠AOC\angle AOC, the middle letter OO is the vertex; the arms point towards AA and CC. Reversing the outside letters names the same marked angle: ∠COA\angle COA. An arc distinguishes the intended angle when a reflex turn is possible.

Angles at a point and on a straight line

Angles that fill one complete turn around a point, without gaps or overlaps, sum to 360∘360^\circ. Adjacent angles filling one side of a straight line sum to 180∘180^\circ.

Not given in the List of formulas: learn these angle facts.

A straight line provides a half-turn because its two rays point in opposite directions. Around a point you need the full turn, including any reflex sector. Do not select two angles merely because they share a vertex; together they must fill the relevant half-turn or full turn.

[DIAGRAM: asset_name: Completing a half-turn or full turn; asset_slug: line_and_point_angle_sums; caption: Identify the whole turn before subtracting the known sectors.; description: Three exact panels show A O B on a straight line with angle AOC 124 degrees and COB unknown; three rays around O partitioning the full turn into 124 degrees, 91 degrees and an unknown sector; and a fresh three-ray full turn partitioned into 137 degrees, 86 degrees and an unknown sector. Every marked arc follows its stated sector.]
Identify the whole turn before subtracting the known sectors.

Worked example 1: one side of a line

Question. A,O,BA,O,B lie on a straight line. Ray OCOC divides one side into ∠AOC=124∘\angle AOC=124^\circ and ∠COB=x\angle COB=x. Find xx and give a reason.

Plan. Subtract the known angle from the half-turn on that side of the line.

x+124∘=180∘angles on a straight linex=180∘−124∘=56∘remove the known sector\begin{aligned} x+124^\circ&=180^\circ && \text{angles on a straight line}\\ x&=180^\circ-124^\circ=56^\circ && \text{remove the known sector} \end{aligned}

Answer. x=56∘x=56^\circ, because angles on a straight line sum to 180∘180^\circ.

Check. 124∘+56∘=180∘124^\circ+56^\circ=180^\circ; an obtuse sector and an acute sector together fill the half-turn.

Now the sectors fill the whole point rather than one side of a straight line. The total changes, not the subtraction strategy.

Worked example 2: the whole point

Question. Three adjacent sectors fill the full turn around OO. Their angles are 124∘124^\circ, 91∘91^\circ and yy. Find yy and give a reason.

Plan. Subtract both known sectors from one full turn.

y=360∘−124∘−91∘angles at a point=145∘calculate the remaining sector\begin{aligned} y&=360^\circ-124^\circ-91^\circ && \text{angles at a point}\\ &=145^\circ && \text{calculate the remaining sector} \end{aligned}

Answer. y=145∘y=145^\circ, because angles at a point sum to 360∘360^\circ.

Check. 124+91+145=360124+91+145=360. Using 180∘180^\circ would give a negative result, warning that a half-turn is the wrong whole.

Watch out

“Angles next to one another add to 180∘180^\circ” needs a straight-line condition. Adjacent angles can total other amounts; use the given straight line or the complete turn.

The fresh check specifies exactly which sectors fill the turn.

Vertically opposite angles

When two straight lines cross, they form two pairs of opposite angles. “Vertically” here means opposite at a vertex; the lines do not need to be vertical on the page.

Vertically opposite angles

Vertically opposite angles are the non-adjacent angles formed by two intersecting straight lines. Each opposite pair is equal.

Why? Each angle in an opposite pair adds to the same neighbouring angle to make 180∘180^\circ. Subtracting that common neighbour leaves equal remainders. This explains the equality without relying on how accurately the picture is drawn.

[DIAGRAM: asset_name: Opposite and neighbouring angles at a crossing; asset_slug: vertically_opposite_crossings; caption: Opposite sectors are equal; neighbouring sectors on a line are supplementary.; description: Three pairs of straight lines AOB and COD cross at O. In the first angle BOC is 68 degrees, in the second 113 degrees and in the third 47 degrees. Angle AOD is labelled x and AOC is labelled y in each. The rays and arcs have exact coordinates consistent with each stated angle.]
Opposite sectors are equal

Worked example 3: acute opposite angles

Question. Straight lines AOBAOB and CODCOD intersect at OO. Given ∠BOC=68∘\angle BOC=68^\circ, find ∠AOD\angle AOD and ∠AOC\angle AOC, giving reasons.

Plan. Copy the angle across the vertex, then complete a straight angle beside it.

∠AOD=68∘vertically opposite angles∠AOC=180∘−68∘=112∘angles on straight line AB\begin{aligned} \angle AOD&=68^\circ && \text{vertically opposite angles}\\ \angle AOC&=180^\circ-68^\circ=112^\circ && \text{angles on straight line }AB \end{aligned}

Answer. ∠AOD=68∘\angle AOD=68^\circ and ∠AOC=112∘\angle AOC=112^\circ.

Check. The other opposite sector is also 112∘112^\circ. Around the point, 68+112+68+112=36068+112+68+112=360.

Continue this lesson.

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