Question

From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones 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 tanαtanβ/tanα+tanβ

Answer 1

$h = \frac{\tan \alpha \tan \beta}{\tan \alpha + \tan \beta}$ (Proved)

Step-by-step explanation:

See the attached diagram.

Distance AC = x miles (say), then distance BC = (1 - x) miles

Now, from the right triangle Δ ACD,

$\tan \alpha = \frac{CD}{AC} = \frac{h}{x}$ {Assumed that CD = height of the plane = h miles}

⇒ $x = \frac{h}{\tan \alpha}$ ............ (1)

Again, from the right triangle Δ BCD,

$\tan \beta = \frac{CD}{BC} = \frac{h}{1 - x}$

⇒ $(1 - x) = \frac{h}{\tan \beta}$ .............. (2)

Now, adding equations (1) and (2) we get,

$x + (1 - x) = \frac{h}{\tan \alpha} + \frac{h}{\tan \beta}$

⇒ $1 = h[\frac{1}{\tan \alpha} + \frac{1}{\tan \beta}]$

⇒ $h = \frac{\tan \alpha \tan \beta}{\tan \alpha + \tan \beta}$ (Proved)

Answer 2

Proved.  Aeroplane is tanαtanβ/tanα+tanβ miles above road.

solution:

• In triangle ABC

tanα =  P/B = AB/BD = h/d

• d = h/tanα_______(1)

•Now, in Triangle ABC

  tanβ = h/(1-d)

• Now, putting d=h/tanα

   tanβ = h/[1- (h/tanα) ]

• now, cross multiplying

 tanβ[1- (h/tanα) ]= h

•  tanβ[ (tanα -h)/tanα) ]= h

• tanβtanα -htanβ = htanα

• tanβtanα = htanα + htanβ

•tanβtanα = h(tanα + tanβ)

• h= (tanβtanα)/(tanα + tanβ)