Question Number 5405 by sanusihammed last updated on 14/May/16
![The angle of elevation of point A and B from P are α and β respectively. The bearing of A and B from P are S20°W and S40°E and their distance from P measured on the map are 3cm and 1cm respetively. A is higher than B. Prove that the elevation of A from B is ((tan^(−1) [3tanα − tanβ])/( (√7) )) Please help me. i have two questions but i have solved one. Please help me to solve this. Thanks for your help.](https://www.tinkutara.com/question/Q5405.png)
$${The}\:{angle}\:{of}\:{elevation}\:{of}\:{point}\:{A}\:{and}\:{B}\:{from}\:{P}\:{are}\:\alpha\:{and}\:\beta\: \\ $$$${respectively}.\:{The}\:{bearing}\:{of}\:{A}\:{and}\:{B}\:{from}\:{P}\:{are}\:{S}\mathrm{20}°{W}\:{and} \\ $$$${S}\mathrm{40}°{E}\:{and}\:{their}\:{distance}\:{from}\:{P}\:\:{measured}\:{on}\:{the}\:{map}\:{are} \\ $$$$\mathrm{3}{cm}\:{and}\:\mathrm{1}{cm}\:{respetively}.\:{A}\:{is}\:{higher}\:{than}\:{B}.\:{Prove}\:{that}\:{the}\: \\ $$$${elevation}\:{of}\:{A}\:{from}\:{B}\:{is} \\ $$$$ \\ $$$$\frac{{tan}^{−\mathrm{1}} \left[\mathrm{3}{tan}\alpha\:−\:{tan}\beta\right]}{\:\sqrt{\mathrm{7}}\:} \\ $$$$ \\ $$$${Please}\:{help}\:{me}.\:{i}\:{have}\:{two}\:{questions}\:{but}\:{i}\:{have}\:{solved}\:{one}. \\ $$$${Please}\:{help}\:{me}\:{to}\:{solve}\:{this}.\:{Thanks}\:{for}\:{your}\:{help}. \\ $$
Commented by Yozzii last updated on 14/May/16
$${The}\:{proof}\:{required}\:{suggests}\:{that},\:{for}\:\theta \\ $$$${being}\:{the}\:{answer}\:{of}\:{angle}\:{of}\:{elevation}, \\ $$$${tan}\left(\theta\sqrt{\mathrm{7}}\right)=\mathrm{3}{tan}\alpha−{tan}\beta. \\ $$$${Is}\:{it}\:{possible}\:{for}\:{you}\:{to}\:{check}\:{the}\:{question} \\ $$$${again}\:{to}\:{make}\:{sure}\:{that}\:{is}\:{what}\:{is} \\ $$$${required}? \\ $$$${Is}\:{it}\:\theta={tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}{tan}\alpha−{tan}\beta}{\:\sqrt{\mathrm{7}}}\right)\:{instead}? \\ $$$$ \\ $$