A plane is observed to be approaching the airport. It is at a...

Let $h$ be the height of the plane above the ground.

The distance from the point of observation to the plane is $d = 10$ km.

The angle of elevation is \theta = 67.5^{}^{\circ}.

From the right-angled triangle formed, $\sin \theta = \dfrac{h}{d}$.

h = 10 \sin 67.5^{}^{\circ}

Using the half-angle formula, $\sin \theta = \sqrt{\dfrac{1 - \cos 2\theta}{2}}$.

\sin 67.5^{}^{\circ} = \sqrt{\dfrac{1 - \cos 135^{}^{\circ}}{2}} = \sqrt{\dfrac{1 - \left(-\dfrac{1}{\sqrt{2}}\right)}{2}} = \sqrt{\dfrac{\sqrt{2} + 1}{2\sqrt{2}}}

Multiplying the numerator and denominator inside the square root by \sqrt{2}:<br><br>\sin 67.5^{}^{\circ} = \sqrt{\dfrac{2 + \sqrt{2}}{4}} = \dfrac{\sqrt{2 + \sqrt{2}}}{2}

Substituting this value into the equation for $h$:

$h = 10 \times \dfrac{\sqrt{2 + \sqrt{2}}}{2} = 5\sqrt{2 + \sqrt{2}}$ km