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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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