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Question-8884




Question Number 8884 by tawakalitu last updated on 03/Nov/16
Answered by sandy_suhendra last updated on 04/Nov/16
Commented by sandy_suhendra last updated on 04/Nov/16
let AL=the horizontal distance pointA to the lower hill  and AH=the horizontal distance pointA to the higher hill  LH=the horizontal distance higher hill to the lower hill  LH=(√(3000^2 +2000^2 −2×3000×2000×cos120°))         =4358.9 m
$$\mathrm{let}\:\mathrm{AL}=\mathrm{the}\:\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{pointA}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{hill} \\ $$$$\mathrm{and}\:\mathrm{AH}=\mathrm{the}\:\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{pointA}\:\mathrm{to}\:\mathrm{the}\:\mathrm{higher}\:\mathrm{hill} \\ $$$$\mathrm{LH}=\mathrm{the}\:\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{higher}\:\mathrm{hill}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{hill} \\ $$$$\mathrm{LH}=\sqrt{\mathrm{3000}^{\mathrm{2}} +\mathrm{2000}^{\mathrm{2}} −\mathrm{2}×\mathrm{3000}×\mathrm{2000}×\mathrm{cos120}°} \\ $$$$\:\:\:\:\:\:\:=\mathrm{4358}.\mathrm{9}\:\mathrm{m} \\ $$
Commented by sandy_suhendra last updated on 04/Nov/16
Commented by sandy_suhendra last updated on 04/Nov/16
y_L =the height of the lower hill      = 3000×tan2°=104.8 m  y_H =the height of the higher hill    = 2000×tan5° =175 meter
$$\mathrm{y}_{\mathrm{L}} =\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{hill} \\ $$$$\:\:\:\:=\:\mathrm{3000}×\mathrm{tan2}°=\mathrm{104}.\mathrm{8}\:\mathrm{m} \\ $$$$\mathrm{y}_{\mathrm{H}} =\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\:\mathrm{higher}\:\mathrm{hill} \\ $$$$\:\:=\:\mathrm{2000}×\mathrm{tan5}°\:=\mathrm{175}\:\mathrm{meter} \\ $$
Commented by sandy_suhendra last updated on 04/Nov/16
Commented by sandy_suhendra last updated on 04/Nov/16
let θ=the elevation angle  tan θ = ((175−104.8)/(4358.9))          θ = 0.92° ≈ 1°
$$\mathrm{let}\:\theta=\mathrm{the}\:\mathrm{elevation}\:\mathrm{angle} \\ $$$$\mathrm{tan}\:\theta\:=\:\frac{\mathrm{175}−\mathrm{104}.\mathrm{8}}{\mathrm{4358}.\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\theta\:=\:\mathrm{0}.\mathrm{92}°\:\approx\:\mathrm{1}° \\ $$
Commented by tawakalitu last updated on 04/Nov/16
i really appreciate sir . God bless you.
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}\:.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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