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In die diagram hieronder is CBFD 'n sirkel zodatning dat BC || FD - NSC Technical Mathematics - Question 8 - 2022 - Paper 2

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In die diagram hieronder is CBFD 'n sirkel zodatning dat BC || FD. CH en DH is raakyne by C en D onderskeidelik. Raaklyne CH en DH sny by H. CF en BD sny by M. $C_A ... show full transcript

Worked Solution & Example Answer:In die diagram hieronder is CBFD 'n sirkel zodatning dat BC || FD - NSC Technical Mathematics - Question 8 - 2022 - Paper 2

Step 1

Bepaal, met redes, die grootte van $H_1$.

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Answer

Die hoeken DA=DB=37extoD_A = D_B = 37^{ ext{o}} omdat hulle uit dieselfde punt kom en die raaklyne aan die sirkel is.

Volgens die tangensstelling het ons: H_1 = 74^{ ext{o}} ext{ (buitelyn $ ot $ in } riangle).

Step 2

Bepaal, met redes, die grootte van $C_2$.

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Answer

CA=DB=37extoC_A = D_B = 37^{ ext{o}} (hoeke in die selfde segment).

C2=F1=37extoC_2 = F_1 = 37^{ ext{o}} (alternatiewe hoeke ext(t.o.v.BCFD) ext{(t.o.v. } BC || FD)).

Step 3

Toon dat MD = MF.

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Answer

Aangesien DA=DB=37extoD_A = D_B = 37^{ ext{o}}, kan ons gebruik maak van die eienskap van hoeke in dieselfde segment.

Ons erken dat: MD = MF ext{ (syne wat teenoor dieselfde $ riangle $ is)}.

Step 4

Bewys dat CHDM 'n koordervierhoek is.

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Answer

In riangle riangle, is H2=180exto74exto=106extoH_2 = 180^{ ext{o}} - 74^{ ext{o}} = 106^{ ext{o}} (som van hoeke in riangle riangle).

H1=HMext(buitelyn)H_1 = H_M ext{ (buitelyn)}

Aangesien CHDM 'n sirkelreël volg, is CHDM 'n koordervierhoek, omdat H1H_1 en H2H_2 aanvullende hoeke (C1C_1 en D1D_1 is aanvullend).

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