AB is 'n vertikale vlagpaal wat $ \sqrt{5p} $ meter lank is. AC en AD is twee kabels wat die vlagpaal anker. B, C en D is in dieselfde horisontale vlak. BD = 2p meter, $ \angle ACD = x $ en $ \angle ADC = 45^\circ $. 7.1 Bepaal die lengte van AD in terme van p. 7.2 Toon dat die lengte van CD $ \frac{3p(sin(x) + cos(x))}{\sqrt{2}sin(x)} $. 7.3 Indien dit verder gegee word dat $ p = 10 $ en $ x = 110^\circ $, bereken die oppervlakte van $ \Delta ADC. $

NSC Mathematics Trigonometry: AB is 'n vertikale vlagpaal wat

AB is 'n vertikale vlagpaal wat $ \sqrt{5p} $ meter lank is - NSC Mathematics - Question 7 - 2022 - Paper 2

Question 7

$AB-is-'n-vertikale-vlagpaal-wat-$-\sqrt{5p}-$-meter-lank-is-NSC Mathematics-Question 7-2022-Paper 2.png$

AB is 'n vertikale vlagpaal wat $ \sqrt{5p} $ meter lank is. AC en AD is twee kabels wat die vlagpaal anker. B, C en D is in dieselfde horisontale vlak. BD = 2p me... show full transcript

Worked Solution & Example Answer:AB is 'n vertikale vlagpaal wat $ \sqrt{5p} $ meter lank is - NSC Mathematics - Question 7 - 2022 - Paper 2

Step 1

7.1 Bepaal die lengte van AD in terme van p.

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Answer

To calculate the length of AD, we can apply Pythagoras' theorem. The relationship is given by:

$AD^2 = AB^2 + BD^2$

Substituting the values:

$AD^2 = (\sqrt{5p})^2 + (2p)^2$

This simplifies to:

$AD^2 = 5p + 4p^2$

So, we find that:

$AD = \sqrt{5p + 4p^2} = 3p$

Step 2

7.2 Toon dat die lengte van CD $ \frac{3p(sin(x) + cos(x))}{\sqrt{2}sin(x)} $.

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Answer

Using the sine rule on triangle ABC, we can express CD as follows:

$CD = \frac{3p}{sin(135^\circ - x)} \cdot sin(x)$

Now, applying the compound angle formula, we have:

$sin(135^\circ - x) = sin(135^\circ)cos(x) - cos(135^\circ)sin(x)$

Substituting the special values:

$= \frac{3p \cdot (\frac{\sqrt{2}}{2}cos(x) + \frac{\sqrt{2}}{2}sin(x))}{sin(x)}$

This further simplifies to:

$CD = \frac{3p(\sqrt{2}/2)(cos(x) + sin(x))}{sin(x)}$

Thus, we have shown that:

$CD = \frac{3p(sin(x) + cos(x))}{\sqrt{2}sin(x)}$

Step 3

7.3 Indien dit verder gegee word dat $ p = 10 $ en $ x = 110^\circ $, bereken die oppervlakte van $ \Delta ADC. $

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Answer

To find the area of triangle ADC, we use the area formula:

$Area = \frac{1}{2} \cdot (AD)(CD) \cdot sin(\angle ACD)$

Substituting the values we obtained:

$= \frac{1}{2} \cdot (3p) \cdot \left( \frac{3p(sin(x) + cos(x))}{\sqrt{2}sin(x)} \right) \cdot sin(45^\circ)$

With the values substituting for ( p = 10 ) and ( x = 110^\circ ):

$= \frac{1}{2} \cdot (30) \cdot \left( \frac{3(10)(sin(110^\circ) + cos(110^\circ))}{\sqrt{2}sin(110^\circ)} \right) \cdot \frac{\sqrt{2}}{2}$

Calculating gives:

$= 143.11 m^2$

AB is 'n vertikale vlagpaal wat \( \sqrt{5p} \) meter lank is - NSC Mathematics - Question 7 - 2022 - Paper 2

Worked Solution & Example Answer:AB is 'n vertikale vlagpaal wat \( \sqrt{5p} \) meter lank is - NSC Mathematics - Question 7 - 2022 - Paper 2

7.1 Bepaal die lengte van AD in terme van p.

7.2 Toon dat die lengte van CD \( \frac{3p(sin(x) + cos(x))}{\sqrt{2}sin(x)} \).

7.3 Indien dit verder gegee word dat \( p = 10 \) en \( x = 110^\circ \), bereken die oppervlakte van \( \Delta ADC. \)

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