In die diagram hieronder is O die middelpunt van sirkel DC. Koord BD word verleng om CF by F te sny sodat $ riangle O_{1} = 4^{ ext{o}} F$ $ heta = 27^{ ext{o}}$ 1. $ heta_{1}$ 2. $ heta_{2}$ 3. $O_{1} = 4 imes 27^{ ext{o}} = 108^{ ext{o}}$ 4. $D_{1} = 54^{ ext{o}}$ 3. Bewys vervolgens dat DC = DF (sides opp. $ riangle E$)

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In die diagram hieronder is O die middelpunt van sirkel DC - NSC Technical Mathematics - Question 7 - 2023 - Paper 2

Question 7

In die diagram hieronder is O die middelpunt van sirkel DC. Koord BD word verleng om CF by F te sny sodat $ riangle O_{1} = 4^{ ext{o}} F$ $ heta = 27^{ ext{o}}$ ... show full transcript

Worked Solution & Example Answer:In die diagram hieronder is O die middelpunt van sirkel DC - NSC Technical Mathematics - Question 7 - 2023 - Paper 2

Step 1

Skryf die grootte van $O_{1}$ neer.

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Answer

$O_{1} = 4 imes 27^{ ext{o}} = 108^{ ext{o}}$

Step 2

Bepaal, met 'n rede, die grootte van $D_{1}$.

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Answer

$D_{1} = 54^{ ext{o}}$
(As die hoeke in die middelpunt = 2 imes ext{hoek by omtrek})

Step 3

Bewys vervolgens dat DC = DF.

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Answer

Gebruik die eienskappe van 'n sirkel: $riangle O_{C} = riangle O_{D}$ = $ext{sides opp.}$ = ext{is emal angles} $\ DC = DF$

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