The-angle-of-elevetion-of-the-top-of-a-tower-from-a-point-A-in-the-north-is-and-the-angle-of-the-top-of-the-tower-from-the-point-B-in-the-east-of-the-point-A-is-prove-that-the-height-of-the-tow

Question Number 177507 by peter frank last updated on 06/Oct/22

The angle of elevetion of the top of a tower from a point A in the north is α and the angle of the top of the tower from the point B in the east of the point A is β .prove that the height of the tower is ((ABsin αsin B)/( (√(sin^2 α−sin^2 β))))

$$\mathrm{The}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevetion}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\: \\ $$$$\mathrm{A}\:\mathrm{in}\:\mathrm{the}\:\mathrm{north}\:\mathrm{is}\:\alpha\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tower} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:\mathrm{B}\:\mathrm{in}\:\mathrm{the}\:\mathrm{east}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{A}\:\mathrm{is}\:\beta\:.\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{height} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{tower}\:\mathrm{is}\: \\ $$$$\frac{\mathrm{ABsin}\:\alpha\mathrm{sin}\:\mathrm{B}}{\:\sqrt{\mathrm{sin}\:^{\mathrm{2}} \alpha−\mathrm{sin}\:^{\mathrm{2}} \beta}} \\ $$

Answered by som(math1967) last updated on 06/Oct/22

let OP=height =x OA=xcotα OB=xcotβ AB=(√(OB^2 −OA^2 )) =(√(x^2 cot^2 β−x^2 cot^2 α)) AB=x(√((cos^2 βsin^2 α−sin^2 βcos^2 α)/(sin^2 αsin^2 β))) sinαsinβAB=x(√(sin^2 α−sin^2 αsin^2 β−sin^2 β+sin^2 αsin^2 β)) ∴x=((ABsinαsinβ)/( (√(sin^2 α−sin^2 β))))

$${let}\:{OP}={height}\:={x} \\ $$$${OA}={xcot}\alpha \\ $$$${OB}={xcot}\beta \\ $$$${AB}=\sqrt{{OB}^{\mathrm{2}} −{OA}^{\mathrm{2}} } \\ $$$$\:\:\:=\sqrt{{x}^{\mathrm{2}} {cot}^{\mathrm{2}} \beta−{x}^{\mathrm{2}} {cot}^{\mathrm{2}} \alpha} \\ $$$${AB}={x}\sqrt{\frac{{cos}^{\mathrm{2}} \beta{sin}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \beta{cos}^{\mathrm{2}} \alpha}{{sin}^{\mathrm{2}} \alpha{sin}^{\mathrm{2}} \beta}} \\ $$$${sin}\alpha{sin}\beta{AB}={x}\sqrt{{sin}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \alpha{sin}^{\mathrm{2}} \beta−{sin}^{\mathrm{2}} \beta+{sin}^{\mathrm{2}} \alpha{sin}^{\mathrm{2}} \beta} \\ $$$$\therefore{x}=\frac{{ABsin}\alpha{sin}\beta}{\:\sqrt{{sin}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \beta}} \\ $$

Commented by som(math1967) last updated on 06/Oct/22

Commented by peter frank last updated on 06/Oct/22

$$\mathrm{thank}\:\mathrm{you} \\ $$

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