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AB represents a vertical netball pole - NSC Mathematics - Question 7 - 2017 - Paper 2
Question 7
AB represents a vertical netball pole. Two players are positioned on either side of the netball pole at points D and E such that D, B and E are on the same straight ... show full transcript
Worked Solution & Example Answer:AB represents a vertical netball pole - NSC Mathematics - Question 7 - 2017 - Paper 2
Step 1
Step 2
Show that AC = \( \frac{k \cdot \tan y}{\sin x} \).
Answer
In triangle ABE:
-
We have:
- AB = k \cdot \tan y
- BE = \tan y
- Therefore, from the right triangle, we know:\n [ AB = k \cdot \tan y ]\n In triangle ABC:
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Applying the sine rule, we find:
- ( \frac{AB}{\sin x} = \frac{AC}{\sin y} )
- Substituting for AB: [ \frac{k \cdot \tan y}{\sin x} = \frac{AC}{\sin y} ]
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Rearranging gives us: [ AC = \frac{k \cdot \tan y}{\sin x} ]
Step 3
If it is further given that ∠DAC = 2x and AD = AC, show that the distance DC between the players at D and C is \( 2k \cdot \tan y \).
Answer
Using the cosine rule in triangle ADC:
- We know that:\n [ DC^2 = AD^2 + AC^2 - 2 \cdot AD \cdot AC \cdot \cos(90° - x) ]\n
- Given that AD = AC = ( \frac{k \cdot \tan y}{\sin x} ):
- Then substituting into the equation gives us: [ DC^2 = 2\left(\frac{k \cdot \tan y}{\sin x}\right)^2 - 2\left(\frac{k \cdot \tan y}{\sin x}\right)^2 \cdot \sin x \cdot \cos 2x]\n
- After simplifying: [ DC = 2k \cdot \tan y ]