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The-angle-of-elevation-of-point-A-and-B-from-P-are-and-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-




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

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