A pilot is flying in a helicopter. At point A, which is h metres directly above point D on the ground, he notices a strange object at point B. The pilot determines that the angle of depression from A to B is 30°. He also determines that the control room at point C is 3h metres from A and ∠BAC = 2x. Points B, C and D are in the same horizontal plane. This scenario is shown in the diagram below. 7.1 Determine the distance AB in terms of h. 7.2 Show that the distance between the strange object at point B and the control room at point C is given by BC = $\frac{h}{25 - 24\cos^2x}$

NSC Mathematics Trigonometry: Sine, Cosine and Area Rules: A pilot is flying in a helicopte

A pilot is flying in a helicopter - NSC Mathematics - Question 7 - 2018 - Paper 2

Question 7

A pilot is flying in a helicopter. At point A, which is h metres directly above point D on the ground, he notices a strange object at point B. The pilot determines t... show full transcript

Worked Solution & Example Answer:A pilot is flying in a helicopter - NSC Mathematics - Question 7 - 2018 - Paper 2

Step 1

7.1 Determine the distance AB in terms of h.

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Answer

To find the distance AB, we need to use the angle of depression from point A to point B, which is 30°. In triangle ABD, we can use the sine function:

$\sin(30°) = \frac{h}{AB}$

Since we know that (\sin(30°) = \frac{1}{2}), we have:

$\frac{1}{2} = \frac{h}{AB}$

Rearranging gives us:

$AB = 2h$

Thus, the distance AB in terms of h is:

$AB = 2h$

Step 2

7.2 Show that the distance between the strange object at point B and the control room at point C is given by BC = $\frac{h}{25 - 24\cos^2x}$

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Answer

To find the distance BC, we apply the cosine rule in triangle ABC, which gives:

$BC^2 = AB^2 + AC^2 - 2AB \cdot AC \cdot \cos(\angle BAC)$

We know:

(AB = 2h)
(AC = 3h)
(\cos(\angle BAC) = \cos(2x))

Substituting these values into the cosine rule, we get:

$BC^2 = (2h)^2 + (3h)^2 - 2(2h)(3h)\cos(2x)$

Calculating further:

$= 4h^2 + 9h^2 - 12h^2 \cos(2x)$ $= 13h^2 - 12h^2 \cos(2x)$

Using the double angle identity, (\cos(2x) = 2\cos^2(x) - 1), we substitute to find:

$BC^2 = 13h^2 - 12h^2(2\cos^2(x) - 1)$ $= 13h^2 - 24h^2\cos^2(x) + 12h^2$ $= 25h^2 - 24h^2\cos^2(x)$

Therefore, we get:

$BC = \sqrt{25h^2 - 24h^2 \cos^2(x)}$

And consequently:

$BC = \frac{h}{25 - 24\cos^2(x)}$

A pilot is flying in a helicopter - NSC Mathematics - Question 7 - 2018 - Paper 2

Worked Solution & Example Answer:A pilot is flying in a helicopter - NSC Mathematics - Question 7 - 2018 - Paper 2

7.1 Determine the distance AB in terms of h.

7.2 Show that the distance between the strange object at point B and the control room at point C is given by BC = \(\frac{h}{25 - 24\cos^2x}\)

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