Question

A tree standing on a horizontal plane is leaning towards the 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)tan alpha tan beta)/(tan alpha - tan beta)`

Answer 1

Let OP be the tree and A, B be the two points such OA = a and OB = b and angle of elevation to the tops are α and β respectively.

Let OL = x and PL = h

We have to prove the following

`h = ((b - a)tan alpha tan beta)/(tan alpha - tan beta)`

The corresponding figure is as follows

In ΔALP

`=> tan alpha = (PL)/(OA + OL)`

`=> tan alpha = h/(a + x)`

`=> 1/(cot alpha) = h/(a + x)`

`=> h cot alpha = a  + x`.....(1)

Again in ΔBLP

`=> tan beta = (PL)/(OB + OL)`

`=> tan beta = h/(b + x)`

`=> 1/(cot beta) = h/(b + x)`

`=> h cot beta  = b + x`    .....(2)

Subtracting equation (1) from (2) we get

`=> h cot beta - hcot alpha = b - a` 

`=> h(cot beta - cot alpha) = b - a`

`=> h (b -a)/(cot beta - cot alpha)`

`h = ((b - a)tan alpha tan beta)/(tan alpha - tan beta)`

Hence height of the top from ground is `h = ((b -a) tan alpha tan beta)/((tan alpha - tan beta))`