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Die kaptein van 'n boot op die see, by punt Q, neem 'n vuring PM direk noord van sy posisie waar - NSC Mathematics - Question 7 - 2018 - Paper 2

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Die kaptein van 'n boot op die see, by punt Q, neem 'n vuring PM direk noord van sy posisie waar. Hy bepaal dat die hoogtehoek van P, die toppunt van die vuring, θ i... show full transcript

Worked Solution & Example Answer:Die kaptein van 'n boot op die see, by punt Q, neem 'n vuring PM direk noord van sy posisie waar - NSC Mathematics - Question 7 - 2018 - Paper 2

Step 1

7.1 Skryf QM in terme van x en θ.

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Answer

In triangle PMQ, applying the tangent function:

tanθ=xQM\tan \theta = \frac{x}{QM}

Rearranging gives:

QM=xtanθQM = \frac{x}{\tan \theta}

We can alternatively express this using the sine rule:

xsinP=QMsinθ\frac{x}{\sin P} = \frac{QM}{\sin \theta}

Hence,

QM=xsinθsinPQM = \frac{x \cdot \sin \theta}{\sin P}

Step 2

7.2 Bewys dat \( \tan \beta = \frac{\cos \beta}{6} \)

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Answer

In triangle PMR, again using the tangent function:

tanβ=MRQM\tan \beta = \frac{MR}{QM}

Where: MR=180°2βMR = 180° - 2\beta and substituting gives:

MR=12xsin(180°2β)12xsinβMR = \frac{12x \cdot \sin(180° - 2\beta)}{12x \cdot \sin \beta}

Simplifying leads to:

tanβ=xcosβ12sinβ\tan \beta = \frac{x \cdot \cos \beta}{12 \cdot \sin \beta}

Substituting the relationships gives:

tanβ6=cosβ\tan \beta \cdot 6 = \cos \beta. Therefore,

tanβ=cosβ6\tan \beta = \frac{\cos \beta}{6}.

Step 3

7.3 Indien β = 40° en QM = 60 meter, bereken die hoogte van die vuring tot die naaste meter.

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Answer

Using the prior relationships:

Given: QM=60QM = 60 meters

you can substitute into:

x=60cos(40°)x = 60 \cdot \cos(40°)

Calculating gives: x45.96x \approx 45.96 meters

To find the height of the lighthouse:

Height=8 meters (given)\text{Height} = 8 \text{ meters (given)}.

Thus, the height would round to 46 meters altogether when considering the height at point P.

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