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AB stel 'n vertikale netbalpaal voor - NSC Mathematics - Question 7 - 2017 - Paper 2

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AB stel 'n vertikale netbalpaal voor. Twee spelers word aan weerskante van die netbalpaal by punt D en E geplaas sodanig dat B, E en D op dieselfde regu lijn is. 'n ... show full transcript

Worked Solution & Example Answer:AB stel 'n vertikale netbalpaal voor - NSC Mathematics - Question 7 - 2017 - Paper 2

Step 1

Skryf die grootte van ABC neer.

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Answer

In triangle ABC, die hoek ABC is regtehoekig, dus:

[ \angle ABC = 90^\circ ]

Step 2

Toon dat AC = \( \frac{k \cdot tan y}{sin x} \).

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Answer

In triangle ABE:

  • AB = k \cdot tan y
  • BE = tan y
  • AB = k \cdot tan y

In triangle ABC:

  • AB = k \cdot tan y
  • AC = sin x

Derde stap:

[ AC = \frac{AB \cdot sin x}{tan y} ]

Substituer met AB: [ AC = \frac{k \cdot tan y \cdot sin x}{tan y} = \frac{k \cdot tan y}{sin x} ]

Step 3

Indien dit verder gegee word dat DAC = 2x en AD = AC = AC, toon dat 2k tan y die afstand DC tussen die spelers by D en C is.

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Answer

Laat ons die sine-reël gebruik:

[ \frac{DC}{sin(90 - x)} = \frac{AC}{sin x} ]

Hieruit volg:

[ DC = \frac{AC \cdot sin(90 - x)}{sin x} ]

Simplifisering: [ DC = \frac{AC \cdot cos x}{sin x} ]

Substitueer AC: [ DC = \frac{k tan y \cdot cos x}{sin x} = 2k tan y ]

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