Die kaptein van 'n boot op die see, by punt Q, neem 'n vaarboom PM direk noord van sy posisie waar - NSC Mathematics - Question 7 - 2018 - Paper 2
Question 7
Die kaptein van 'n boot op die see, by punt Q, neem 'n vaarboom PM direk noord van sy posisie waar. Hy bepaal dat die hoogtehoek van P, die toppunt van die vaarboom,... show full transcript
Worked Solution & Example Answer:Die kaptein van 'n boot op die see, by punt Q, neem 'n vaarboom PM direk noord van sy posisie waar - NSC Mathematics - Question 7 - 2018 - Paper 2
Step 1
Skryf QM in terme van x en θ.
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Answer
In triangle PMQ:
Using the definition of tangent in right triangle,
tan(θ)=QMx
Therefore,
QM=tan(θ)x
This gives us the expression for QM in terms of x and θ.
Step 2
Bewys dat $ tan \beta = \frac{6}{12x} $.
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Answer
In triangle PMR:
Using the tangent definition, we have:
tan(β)=QMMR
Here, MR can be calculated:
MR=180°−2β
Substituting into the expression:
QM=6
Thus,
tan(β)=12x6
This proves the relationship for tan(β).
Step 3
Indien $ \beta = 40° $ en $ QM = 60 meter $, bereken die hoogte van die vulkane tot die naaste meter.
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Answer
From the earlier relation:
tan(40°)=60x
Calculating x gives:
x=60⋅tan(40°)
Approximating:
x=60⋅0.8391=50.35
Thus, the height is:
Height=7−x=7−50≈43
Rounding off gives us 43 meters.
