A-person-observes-the-angle-of-elevation-of-the-peak-of-a-hill-from-a-station-to-be-He-walks-c-metres-along-a-slope-inclined-at-the-angle-and-finds-the-angle-of-elevation-of-the-peak-of-the-hill-

Question Number 20021 by Tinkutara last updated on 20/Aug/17

A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at the angle β and finds the angle of elevation of the peak of the hill to be γ. Show that the height of the peak above the ground is ((c sin α sin (γ − β))/((sin γ − α))).

$$\mathrm{A}\:\mathrm{person}\:\mathrm{observes}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{a}\:\mathrm{hill}\:\mathrm{from}\:\mathrm{a}\:\mathrm{station}\:\mathrm{to}\:\mathrm{be} \\ $$$$\alpha.\:\mathrm{He}\:\mathrm{walks}\:{c}\:\mathrm{metres}\:\mathrm{along}\:\mathrm{a}\:\mathrm{slope} \\ $$$$\mathrm{inclined}\:\mathrm{at}\:\mathrm{the}\:\mathrm{angle}\:\beta\:\mathrm{and}\:\mathrm{finds}\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{hill}\:\mathrm{to}\:\mathrm{be}\:\gamma.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{peak}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{is} \\ $$$$\frac{{c}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\left(\gamma\:−\:\beta\right)}{\left(\mathrm{sin}\:\gamma\:−\:\alpha\right)}. \\ $$

Answered by ajfour last updated on 20/Aug/17

Commented by ajfour last updated on 21/Aug/17

hcot α=ccos β+(h−csin β)cot γ ⇒h(((cos α)/(sin α))−((cos γ)/(sin γ)))=c(cos β−((sin βcos γ)/(sin γ))) ((hsin (γ−α))/(sin αsin γ))=((csin (γ−β))/(sin γ)) ⇒ h=((c sin (𝛄−𝛃)sin 𝛂)/(sin (𝛄−𝛂))) .

$$\:\:{h}\mathrm{cot}\:\alpha={c}\mathrm{cos}\:\beta+\left({h}−{c}\mathrm{sin}\:\beta\right)\mathrm{cot}\:\gamma \\ $$$$\Rightarrow{h}\left(\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}−\frac{\mathrm{cos}\:\gamma}{\mathrm{sin}\:\gamma}\right)={c}\left(\mathrm{cos}\:\beta−\frac{\mathrm{sin}\:\beta\mathrm{cos}\:\gamma}{\mathrm{sin}\:\gamma}\right) \\ $$$$\frac{{h}\mathrm{sin}\:\left(\gamma−\alpha\right)}{\mathrm{sin}\:\alpha\mathrm{sin}\:\gamma}=\frac{{c}\mathrm{sin}\:\left(\gamma−\beta\right)}{\mathrm{sin}\:\gamma}\: \\ $$$$\Rightarrow\:\:\:\boldsymbol{{h}}=\frac{\boldsymbol{{c}}\:\mathrm{sin}\:\left(\boldsymbol{\gamma}−\boldsymbol{\beta}\right)\mathrm{sin}\:\boldsymbol{\alpha}}{\mathrm{sin}\:\left(\boldsymbol{\gamma}−\boldsymbol{\alpha}\right)}\:. \\ $$

Commented by Tinkutara last updated on 21/Aug/17

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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