Question

A tower subtends an angle alpha at a point A in the plane of its base and angle of depression of the foot of the tower at a point "b" metres just above A is Beta. porve that the height of the tower is 'b tan alpha cot beta'??????

Answer 1

let AB = b, ED= h (height), AD=BC= x
in triangle BAD,
Perpendicular/ base = tan beeta
= b/x= tan beeta
or
x= b/ tan beeta -----(1)

in triangle AED,
BASE/PERPENDICULAR =cot alpha
= x/h = cot alpha
or
x = hcot alpha -------(2)
putting value of x in (1) we get,
hcot alpha = b/tan beeta
h= b/ cot alpha tan beeta
as (1/cot theeta = tan theeta)
therefore, ( reciprocal) =
h = btan alpha cot beeta

hence proved.
hope it helps.

Answer 2

Answer:

Step-by-step explanation:

let AB = b, ED= h (height), AD=BC= x  

in triangle BAD,

Perpendicular/ base = tan beeta  

= b/x= tan beeta

or  

x= b/ tan beeta -----(1)

in triangle AED,

BASE/PERPENDICULAR =cot alpha

= x/h = cot alpha

or

x = hcot alpha -------(2)  

putting value of x in (1) we get,

hcot alpha = b/tan beeta

h= b/ cot alpha tan beeta  

as (1/cot theeta = tan theeta)

therefore, ( reciprocal) =  

h = btan alpha cot beeta