In die diagram hieronder is CBFD 'n sirkel sodanig dat BCI|FD - NSC Technical Mathematics - Question 8 - 2022 - Paper 2
Question 8
In die diagram hieronder is CBFD 'n sirkel sodanig dat BCI|FD.
CH en DH is raaklynse by C en D onderskeidelik. Raaklyne CH en DH sny by H.
CF en BD sny by M.
Câ = 37... show full transcript
Worked Solution & Example Answer:In die diagram hieronder is CBFD 'n sirkel sodanig dat BCI|FD - NSC Technical Mathematics - Question 8 - 2022 - Paper 2
Step 1
8.1 Bepaal, met redes, die grootte van H₁.
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Answer
Gegee dat Câ = 37°.
Hier gebruik ons die eienskap dat die raaklyne wat van die selfde punt kom 'n gelyke hoek met die koord vorm.
Dus is H₁ = 74° (eksterieur/buite ⊥ in Δ).
Step 2
8.2 Bepaal, met redes, die grootte van C₂.
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Answer
Omdat C₂ = F₁ = 37° (tan-chord th./r.klyn-koord st),
gevolglik is C₂ = 37° (alternatief verw. ⊥; BC|FD).
Step 3
8.3 Toon dat MD = MF.
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Answer
Aangesien Ĉ₂ = D̂₂ = 37° (∠s in die selfde segment),
F₁ = 37° (proef/bewys in 8.2),
dit lei tot MD = MF (sye opp/sye teenoor = ∠).
Step 4
8.4 Bewys dat CHDM 'n koordvierhoek is.
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Answer
Aangesien H₂ = 180° − 74° = 106° (sommige ⊿s ∠s van Δ),
H₁ = M₁ (eksterieur/buite ⊥ in Δ).
CHDM is 'n sirkel vierhoek (ops. as opp pint C).
Dit volg dat CHDM 'n koordvierhoek is omdat C₂ en H₁ supplemeir. (∠s = 180°).
