The angle of elevation of the top of a tower from a point A due South of the tower is - and from B due East of the tower is - If AB - d- show that the height of the tower is d-cot2 - - cot2 -

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Mathematics

Heights and Distances

The angle of ...

Question

The angle of elevation of the top of a tower from a point A due South of the tower is $α$ and from B due East of the tower is $β$ . If AB = d, show that the height of the tower is
$\frac{d}{\sqrt{(c o t^{2} α + c o t^{2} β)}}$

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Solution

PQ is $⊥$ to PA and PB both
$h = P A t a n α$ and $h = P B t a n β$
$∴ P A = h c o t α, P B = h c o t β$ .
Also, $A B^{2} = P A^{2} + P B^{2}$
or $d^{2} = h^{2} (c o t^{2} α + c o t^{2} β)$ .
$∴ h = \frac{d}{\sqrt{c o t^{2} α + c o t^{2} β}}$

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The angle of elevation of the top of a tower from a point A due South of the tower is α and from B due East of the tower is β. If AB = d, show that the height of the tower isd√(cot2α+cot2β)

PQ is ⊥ to PA and PB bothh=PAtanα and h=PBtanβ∴PA=hcotα,PB=hcotβ.Also, AB2=PA2+PB2or d2=h2(cot2α+cot2β).∴h=d√cot2α+cot2β

The angle of elevation of the top of a tower from a point A due South of the tower is $α$ and from B due East of the tower is $β$ . If AB = d, show that the height of the tower is
$\frac{d}{\sqrt{(c o t^{2} α + c o t^{2} β)}}$

PQ is $⊥$ to PA and PB both
$h = P A t a n α$ and $h = P B t a n β$
$∴ P A = h c o t α, P B = h c o t β$ .
Also, $A B^{2} = P A^{2} + P B^{2}$
or $d^{2} = h^{2} (c o t^{2} α + c o t^{2} β)$ .
$∴ h = \frac{d}{\sqrt{c o t^{2} α + c o t^{2} β}}$