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In die diagram lê S, T en K in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2023 - Paper 2

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In die diagram lê S, T en K in dieselfde horizontale vlak. RS is 'n vertikale toring. Die dieptehoek van R na K is β. TSK = α, TS = p meter en die oppervlak van ΔSTK... show full transcript

Worked Solution & Example Answer:In die diagram lê S, T en K in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2023 - Paper 2

Step 1

7.1 Bepaal die lengte van SK in terme van p, q en a.

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Answer

Die area van ΔSTK kan gegee word as: q=12p(SK)sin⁡αq = \frac{1}{2} p (SK) \sin α Hieruit volg: SK=2qpsin⁡αSK = \frac{2q}{p \sin α}

Step 2

7.2 Toon dat RS = \( \frac{2q \tan β}{p \sin α} \)

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Answer

Die trigonometriese verhouding kan geformuleer word as: RKSRS=tan⁡β\frac{RKS}{RS} = \tan β. Hieruit kan ons skryf: RS=SKtan⁡βRS = \frac{SK}{\tan β}
Substitusie gee ons dus: RS=2qpsin⁡α⋅1tan⁡β=2qtan⁡βpsin⁡αRS = \frac{2q}{p \sin α} \cdot \frac{1}{\tan β} = \frac{2q \tan β}{p \sin α}

Step 3

7.3 Bereken die grootte van α as \( α < 90° \) een RS = 70 m, p = 80 m, q = 2 500 m² en β = 42°.

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Answer

Begin met die substitusie in die formule: RS=2qtan⁡42°psin⁡αRS = \frac{2q \tan 42°}{p \sin α} 70=2(2500)tan⁡42°80sin⁡α70 = \frac{2(2500) \tan 42°}{80 \sin α} Los die vergelyking op om α te vind: sin⁡α=2(2500)tan⁡42°70⋅80\sin α = \frac{2(2500) \tan 42°}{70 \cdot 80} Bereken die waarde: sin⁡α=0.28\sin α = 0.28 Gevolglik, α ≈ 53.51°.

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