Question

If the angle of depression and elevation of the top of a tower of height h from the top and bottom of a second tower are x and y respectively, then the height of the second tower is

Answer 1

Answer:

Step-by-step explanati

Answer 2

Answer:

Height of Tower 2 is $h(1+\frac{tan\,x}{tan\,y})$

Step-by-step explanation:

Given: Angle of elevation from botton of tower 2 to top of tower 1 = y

           Angle of depression from top of tower 2 to top of tower 1 = x

To find: Height of tower tower 2.

We know that angle of depression from tower 2 is equal to angle of elevation from tower 1 to top of tower 2.

In ΔBDC,

$tan\,y=\frac{CB}{CD}$

$tan\,y=\frac{h}{CD}$

$CD=\frac{h}{tan\,y}$

In ΔABE,

$tan\,x=\frac{AE}{BE}$

$tan\,x=\frac{AE}{CD}$   ( BE = CD )

$tan\,x=\frac{AE}{\frac{h}{tan\,y}}$

$\frac{h\times tan\,x}{tan\,y}=AE$

Height of tower 2 = $AE+DE=h+\frac{h\times tan\,x}{tan\,y}=h(1+\frac{tan\,x}{tan\,y})$

Therefore, Height of Tower 2 is $h(1+\frac{tan\,x}{tan\,y})$