Question

A man observes the angle of elevation of the top of a building to be 30° . He walks towards it in a horizontal line through its base . On covering 60metres the angle of elevation changes to 60° . Find the height of the building to the nearest metre​

Answer 1

Solution:

Given:

=> Angle of elevation from point A is 30°

=>  Angle of elevation from point D is 60°

=> Distance of DC = 60 m

To Find:

=> Height of building.

So, Let BD = x and AB = h.

Now, In ΔABD

$\sf{\implies \tan 60^{\circ}=\dfrac{AB}{BD}}$

$\sf{\implies \sqrt{3} =\dfrac{h}{x}}$

$\sf{\implies x = \dfrac{h}{\sqrt{3}}\;\;\;\;......(1)}$

Now, in ΔABC

$\sf{\implies \tan 30^{\circ}=\dfrac{AB}{BC}}$

$\sf{\implies \dfrac{1}{\sqrt{3}}=\dfrac{h}{x+60}}$

$\sf{\implies x+60=\sqrt{3} h}$

Now put the value of x from Equation (1), we get

$\sf{\implies \dfrac{h}{\sqrt{3}} +60=\sqrt{3} h}$

$\sf{\implies 60 = \sqrt{3}h-\dfrac{h}{\sqrt{3}}}$

$\sf{\implies 60=\dfrac{3h-h}{\sqrt{3}}}$

$\sf{\implies 2h=60\sqrt{3}}$

$\sf{\implies h = 30\sqrt{2}}$

We know that value of $\sf{\sqrt{3}}$ is 1.73. So,

$\sf{\implies h = 30\times 1.73}$

$\sf{\implies h = 51.9\;m}$

Hence, height of the building is 51.9 m.