Question

A balloon moving in a straight line passes vertically above two points $A$ and $B$ on a horizontal plane $1000ft.$ apart. When above $A$ it has an altitude of ${60}^o$ an seen from $B$ and when above $B$ it has an altitude of ${45}^o$ as seen from $A$. Find the distance from $A$ of the point at which it will touch the plane.

Answer 1

Let the balloon touch the horizontal plane through $A$ in $R$. We have to determine $AR$.
$AB=1000$ and let $BR=x$.

In $△PAB$
$PA=AB\tan {60}^o=1000\sqrt{3}$

In $△QAB$,

$\angle QAB=\angle BQA$

$\therefore AB=QB=1000$

Now from similar triangles,$△APR$ and $△BQR$, we have
$\tan \theta =\frac{AP}{AR}=\frac{BQ}{BR}⟹\frac{1000\sqrt{3}}{1000+x}=\frac{1000}{x}$
$⟹x\sqrt{3}=1000+x$

$⟹x(\sqrt{3}-1)=1000$
$⟹x=\frac{1000}{\sqrt{\left(3\right)}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{1000(\sqrt{3}+1)}{3-1}=500(\sqrt{3}+1)$

$\therefore AR=1000+x$

$=1000+500(\sqrt{3}+1)$

$=500(2+\sqrt{3}+1)=500\sqrt{3}(\sqrt{3}+1)$ ft.