Question Number 20021 by Tinkutara last updated on 20/Aug/17
$$\mathrm{A}\:\mathrm{person}\:\mathrm{observes}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{a}\:\mathrm{hill}\:\mathrm{from}\:\mathrm{a}\:\mathrm{station}\:\mathrm{to}\:\mathrm{be} \\ $$$$\alpha.\:\mathrm{He}\:\mathrm{walks}\:{c}\:\mathrm{metres}\:\mathrm{along}\:\mathrm{a}\:\mathrm{slope} \\ $$$$\mathrm{inclined}\:\mathrm{at}\:\mathrm{the}\:\mathrm{angle}\:\beta\:\mathrm{and}\:\mathrm{finds}\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{hill}\:\mathrm{to}\:\mathrm{be}\:\gamma.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{peak}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{is} \\ $$$$\frac{{c}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\left(\gamma\:−\:\beta\right)}{\left(\mathrm{sin}\:\gamma\:−\:\alpha\right)}. \\ $$
Answered by ajfour last updated on 20/Aug/17
Commented by ajfour last updated on 21/Aug/17
$$\:\:{h}\mathrm{cot}\:\alpha={c}\mathrm{cos}\:\beta+\left({h}−{c}\mathrm{sin}\:\beta\right)\mathrm{cot}\:\gamma \\ $$$$\Rightarrow{h}\left(\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}−\frac{\mathrm{cos}\:\gamma}{\mathrm{sin}\:\gamma}\right)={c}\left(\mathrm{cos}\:\beta−\frac{\mathrm{sin}\:\beta\mathrm{cos}\:\gamma}{\mathrm{sin}\:\gamma}\right) \\ $$$$\frac{{h}\mathrm{sin}\:\left(\gamma−\alpha\right)}{\mathrm{sin}\:\alpha\mathrm{sin}\:\gamma}=\frac{{c}\mathrm{sin}\:\left(\gamma−\beta\right)}{\mathrm{sin}\:\gamma}\: \\ $$$$\Rightarrow\:\:\:\boldsymbol{{h}}=\frac{\boldsymbol{{c}}\:\mathrm{sin}\:\left(\boldsymbol{\gamma}−\boldsymbol{\beta}\right)\mathrm{sin}\:\boldsymbol{\alpha}}{\mathrm{sin}\:\left(\boldsymbol{\gamma}−\boldsymbol{\alpha}\right)}\:. \\ $$
Commented by Tinkutara last updated on 21/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$