A balloon moving in a straight line passes vertically above two...

In $∆ ABE$,
$tan {45}^{o} = \dfrac{h}{10} ⇒ h = 10ft \ldots \ldots \ldots . . \left(i\right)<br />And$ in ∆ ABD,$<br />$tan {60}^{o} = $\dfrac{H}{10}⇒ H =$10\sqrt{3}$ft \ldots \ldots \left(ii\right)<br />Also, \dfrac{H}{10 + x} = \dfrac{h}{BC} ( Similar triangle property)<br />Using equation$\left(i\right)$and \left( ii\right), we get,</p><p>\dfrac{10\sqrt{3}}{10 + x} = \dfrac{10}{x}<br />$⇒ $\sqrt{3}$ x = 10 + x$<br />$⇒ x = $\dfrac{10}{\sqrt{3} - 1} =5 \left(\sqrt{3} +1\right)ft$