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 θ – a tan θ tan α
h tan α = tan θ (h – a 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