Question Number 5405 by sanusihammed last updated on 14/May/16
$${The}\:{angle}\:{of}\:{elevation}\:{of}\:{point}\:{A}\:{and}\:{B}\:{from}\:{P}\:{are}\:\alpha\:{and}\:\beta\: \\ $$$${respectively}.\:{The}\:{bearing}\:{of}\:{A}\:{and}\:{B}\:{from}\:{P}\:{are}\:{S}\mathrm{20}°{W}\:{and} \\ $$$${S}\mathrm{40}°{E}\:{and}\:{their}\:{distance}\:{from}\:{P}\:\:{measured}\:{on}\:{the}\:{map}\:{are} \\ $$$$\mathrm{3}{cm}\:{and}\:\mathrm{1}{cm}\:{respetively}.\:{A}\:{is}\:{higher}\:{than}\:{B}.\:{Prove}\:{that}\:{the}\: \\ $$$${elevation}\:{of}\:{A}\:{from}\:{B}\:{is} \\ $$$$ \\ $$$$\frac{{tan}^{−\mathrm{1}} \left[\mathrm{3}{tan}\alpha\:−\:{tan}\beta\right]}{\:\sqrt{\mathrm{7}}\:} \\ $$$$ \\ $$$${Please}\:{help}\:{me}.\:{i}\:{have}\:{two}\:{questions}\:{but}\:{i}\:{have}\:{solved}\:{one}. \\ $$$${Please}\:{help}\:{me}\:{to}\:{solve}\:{this}.\:{Thanks}\:{for}\:{your}\:{help}. \\ $$
Commented by Yozzii last updated on 14/May/16
$${The}\:{proof}\:{required}\:{suggests}\:{that},\:{for}\:\theta \\ $$$${being}\:{the}\:{answer}\:{of}\:{angle}\:{of}\:{elevation}, \\ $$$${tan}\left(\theta\sqrt{\mathrm{7}}\right)=\mathrm{3}{tan}\alpha−{tan}\beta. \\ $$$${Is}\:{it}\:{possible}\:{for}\:{you}\:{to}\:{check}\:{the}\:{question} \\ $$$${again}\:{to}\:{make}\:{sure}\:{that}\:{is}\:{what}\:{is} \\ $$$${required}? \\ $$$${Is}\:{it}\:\theta={tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}{tan}\alpha−{tan}\beta}{\:\sqrt{\mathrm{7}}}\right)\:{instead}? \\ $$$$ \\ $$