From an aeroplane vertically over a straight horizontal road, the angles of depression of two consecutive milestones on opposite sides of the aeroplane are observed to be $α$ and $β$ . Show that the height in miles of aeroplane above the road is given by
$\frac{tan α tan β}{tan α + tan β} .$

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Solution

$h = B D tan α = D C tan β$
But $B D + D C = 1 m i l e$
$h (\frac{1}{tan α} + \frac{1}{tan β}) = 1$
$∴ h = \frac{tan α tan β}{tan α tan β}$

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From an aeroplane vertically over a straight horizontal road, the angles of depression of two consecutive milestones on opposite sides of the aeroplane are observed to be α and β. Show that the height in miles of aeroplane above the road is given bytanαtanβtanα+tanβ.

h=BDtanα=DCtanβBut BD+DC=1mileh(1tanα+1tanβ)=1∴h=tanαtanβtanαtanβ

From an aeroplane vertically over a straight horizontal road, the angles of depression of two consecutive milestones on opposite sides of the aeroplane are observed to be $α$ and $β$ . Show that the height in miles of aeroplane above the road is given by
$\frac{tan α tan β}{tan α + tan β} .$

$h = B D tan α = D C tan β$
But $B D + D C = 1 m i l e$
$h (\frac{1}{tan α} + \frac{1}{tan β}) = 1$
$∴ h = \frac{tan α tan β}{tan α tan β}$