Question

While standing a man sees the top of a building and founds that the angle of elevation is $\theta$.When he moves x m far from that point, then he founds that the angle of elevation is $\phi$.Then prove that height of the building is $\dfrac{x tan\theta . tan\phi}{tan\theta - tan\phi}$​

Answer 1

Step-by-step explanation:

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}$

For figure refer to attachment:

Let us take that the height of the tower is h.

and let BC= y.

Now, in ∆ABD,

$tan\theta =\dfrac{AB}{BD}=\dfrac{h}{y}$

Therefore, h =y$tan\theta$ .........(1)

Again in ∆ABC,

$tan\phi=\dfrac{AB}{BC}=\dfrac{h}{x+y}$

=> $(x+y) tan\phi = h$

=>$tan\phi(x + \dfrac{h}{tan\theta} ) = h$

$\large\red{\boxed{Since\:\: h = ytan\theta (just proved)}}$

On opening the brackets, we have,

=>$xtan\phi +\dfrac{htan\theta}{tan\theta} = h$

=> $\dfrac{xtan\phi.tan\theta + htan\phi}{tan\theta} = h$

=>$xtan\phi tan\theta+ htan\phi = htan\theta$

=>$xtan\phi tan\theta = htan\theta- htan\phi$

=>$htan\theta- htan\phi = xtan\phi tan\theta$

=>$h( tan\theta - tan\phi)=xtan\phi tan\theta$

=> $h = \dfrac{x tan\theta . tan\phi}{tan\theta - tan\phi}$

$\large\purple{\boxed{Hence\:proved\:,h = \dfrac{x tan\theta . tan\phi}{tan\theta - tan\phi}}}$