'n Landskapskunstenaar beplan om blomme binne twee konsentiese sikrels rondom 'n vertikale lamppaal PQ te plant. R is 'n punt op die binerste sirkel en S is 'n punt op die buitensirkel. R, Q en S lê op dieselfde horizontale vlak. RS is 'n pyp wat vir die besproeiingstelsel van die tuin gebruik word. - Die radius van die binerste sirkel is r eenhede en die radius van die buitensirkel is QS. - Die hoogtevlak vanaf S na P is 30º. - RQS = 2x en PQ = √3r.

NSC Mathematics Trigonometry: 'n Landskapskunstenaar beplan om

'n Landskapskunstenaar beplan om blomme binne twee konsentiese sikrels rondom 'n vertikale lamppaal PQ te plant - NSC Mathematics - Question 7 - 2020 - Paper 2

Question 7

'n Landskapskunstenaar beplan om blomme binne twee konsentiese sikrels rondom 'n vertikale lamppaal PQ te plant. R is 'n punt op die binerste sirkel en S is 'n punt ... show full transcript

Worked Solution & Example Answer:'n Landskapskunstenaar beplan om blomme binne twee konsentiese sikrels rondom 'n vertikale lamppaal PQ te plant - NSC Mathematics - Question 7 - 2020 - Paper 2

Step 1

7.1 Toon dat QS = 3r

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Answer

To prove that QS = 3r, we can use the tangent ratio from the right triangle formed by points P, S, and Q.

an(30^ ext{o}) = \frac{QS}{ ext{height}} = \frac{QS}{\sqrt{3r}}.

From the trigonometric ratio, we also know that:

QS = \sqrt{3r} \tan(30^ ext{o}).

Substituting for \tan(30^ ext{o}) = \frac{1}{\sqrt{3}}:

QS = \sqrt{3r} \times \frac{1}{\sqrt{3}} = r.

Thus,

QS = 3r.

Step 2

7.2 Bepaal, in terme van r, die oppervlakte van die blomtuin.

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Answer

The area of the flower garden can be found by subtracting the area of the inner circle from the area of the outer circle.

Area of outer circle (radius QS = 3r):

A_{outer} = \pi (3r)^{2} = 9\pi r^{2}.

Area of inner circle (radius r):

A_{inner} = \pi r^{2}.

Therefore, the area of the flower garden:

A = A_{outer} - A_{inner} = 9\pi r^{2} - \pi r^{2} = 8\pi r^{2}.

Step 3

7.3 Toon aan dat RS = r/10 - 6 cos 2x.

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Answer

To prove that RS = r/10 - 6 cos 2x, we use the cosine rule in triangle PSR:

RS^{2} = r^{2} + (3r)^{2} - 2(r)(3r)\cos(2x).

Expanding this gives:

RS^{2} = r^{2} + 9r^{2} - 6r^{2}\cos(2x),

which simplifies to:

RS^{2} = (r^{2}(10 - 6\cos(2x))).

Taking the square root:

RS = \frac{r}{10 - 6\cos(2x)}.

Step 4

7.4 Indien r = 10 meter en x = 56º, bereken RS.

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Answer

Substituting r = 10 meters and x = 56º:

Convert 56º to radians:

RS = \frac{10}{10 - 6\cos(56^ ext{o})}.

Calculating \cos(56^ ext{o}):

RS = \frac{10}{10 - 6(0.559)} = \frac{10}{10 - 3.354} = \frac{10}{6.646} \approx 1.505.

Thus, RS ≈ 1.51 meters.

'n Landskapskunstenaar beplan om blomme binne twee konsentiese sikrels rondom 'n vertikale lamppaal PQ te plant - NSC Mathematics - Question 7 - 2020 - Paper 2

Worked Solution & Example Answer:'n Landskapskunstenaar beplan om blomme binne twee konsentiese sikrels rondom 'n vertikale lamppaal PQ te plant - NSC Mathematics - Question 7 - 2020 - Paper 2

7.1 Toon dat QS = 3r

7.2 Bepaal, in terme van r, die oppervlakte van die blomtuin.

7.3 Toon aan dat RS = r/10 - 6 cos 2x.

7.4 Indien r = 10 meter en x = 56º, bereken RS.

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