A tree standing on a horizontal plane is leaning towards east. At two points situated at distances a and b exactly due west on it , the angles of elevation of the top are respectively α and β. Prove that the height of the top from the ground is  (b-a)tanαtanβ/tanα-tanβ.

 

Let PQ be the leaning tree and R and S be the two given points at distance a and b from P respectively.

Let PT = x and QT = h

 

In ∆PQT

 

In ∆QRT,

 

Putting the value of x from (1) in (2), we get

⇒ tan α (h + a tan θ) = h tan θ

h tan α + a tan θ tan α = h tan θ

h  tan α = h tan θa tan θ tan α

h tan α = tan θ (ha tan α)

 

Now, In ∆QST

 

Substituting the value of tan θ from (3), we get

h2 tan β – ah tan α tan β + bh tan α tan β – h2 tan α = 0

h (h tan β – h tan α + b tan α tan β – a tan α tan β) = 0

h (tan β – tan α) + tan α tan β (b a) = 0

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