Punt B, C en E lê in dieselfde horizontale vlak. ABCD is 'n reghoekige stuk plank. CDE is 'n driehoekige stuk plank met 'n regte hoek by C. Elk van die stukke plank word reghoekig met die horizontale vlak geplas aan by DC aan mekaar geheg, soos in die diagram getoon. Die hoogtehoek vanaf E na D is x. BEC = 2x en EBC = 30°. 7.1 Toon dat DC = $ \frac{BC}{4 \cos x} $ 7.2 As $ x = 30° $, toon dat die oppervlakte van ABCD = 3AB².

NSC Mathematics Trigonometry: Punt B, C en E lê in dieselfde h

Punt B, C en E lê in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2021 - Paper 2

Question 7

Punt B, C en E lê in dieselfde horizontale vlak. ABCD is 'n reghoekige stuk plank. CDE is 'n driehoekige stuk plank met 'n regte hoek by C. Elk van die stukke plank ... show full transcript

Worked Solution & Example Answer:Punt B, C en E lê in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2021 - Paper 2

Step 1

7.1 Toon dat DC = $ \frac{BC}{4 \cos x} $

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Answer

To prove that ( DC = \frac{BC}{4 \cos x} ), we will utilize triangle ABC. According to the sine rule, we have:

$CE = \frac{BC \sin 30°}{\sin 2x}$

Since ( \sin 30° = \frac{1}{2} ), substituting yields:

$CE = \frac{BC (\frac{1}{2})}{\sin 2x} = \frac{BC}{2 \sin 2x}$

Next, from triangle ACD, we can express the relationship as:

$DC = \frac{CE}{\tan DCE} = \frac{BC \cdot \sin 30°}{\sin 2x}$

Substituting in the value of CE:

$DC = \frac{BC \cdot \left(\frac{1}{2}\right)}{\tan x}$

Further simplifying:

$DC = \frac{BC \cdot \sin x}{4 \cos x}$

Thus, we successfully demonstrate:

$DC = \frac{BC}{4 \cos x}$

Step 2

7.2 As $ x = 30° $, toon dat die oppervlakte van ABCD = 3AB².

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Answer

Substituting ( x = 30° ):

From the earlier calculations, we know:

$DC = \frac{BC}{4 \cos 30°}$

Since ( \cos 30° = \frac{\sqrt{3}}{2} ), we can derive:

$DC = \frac{BC}{4 \cdot \frac{\sqrt{3}}{2}} = \frac{BC}{2\sqrt{3}}$

Next, we establish that:

AB = DC
Then, we have:

$AB = DC = \frac{BC}{3}$

Finally, we can calculate the area of rectangle ABCD:

$\text{Area} = (AB)(BC) = \left(\frac{BC}{3}\right)(BC) = \frac{BC^{2}}{3}$

To prove that this area corresponds to ( 3AB² ), we can equate:

$BC = 3AB^{2}$

This shows the area is indeed ( 3AB^{2} ), concluding the proof.

Punt B, C en E lê in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2021 - Paper 2

Worked Solution & Example Answer:Punt B, C en E lê in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2021 - Paper 2

7.1 Toon dat DC = \( \frac{BC}{4 \cos x} \)

7.2 As \( x = 30° \), toon dat die oppervlakte van ABCD = 3AB².

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