Solveeit Logo

Question

Question: Find the height of a building, when it is found that on walking toward it \( 40{\rm{ m}} \) in a hor...

Find the height of a building, when it is found that on walking toward it 40m40{\rm{ m}} in a horizontal line through its base the angular elevation of its top changes from 3030^\circ to 4545^\circ .

Explanation

Solution

Hint : This question is based on Trigonometry. In this question we have to find the height of a building using the trigonometric relations. To solve this question, we first draw the diagram for this problem showing the height of the building and the angles that the person walking towards the building makes with the top of the building. Then using the trigonometric formula, we can calculate the height of the building.

Complete step-by-step answer :
Given:
Let us assume the height AB of the building is hh metres.
Also, assuming the person is initially at point D on the ground and from this point the angle of the elevation to the top of the building is 3030^\circ .
Now after moving CD=40mCD = 40{\rm{ m}} towards the building the person reaches at point C on the ground and from this point the angle of the top of the building is 4545^\circ .

In ΔABC\Delta ABC we have,
tan45=ABBC\tan 45^\circ = \dfrac{{AB}}{{BC}}
Since the value of tan45=1\tan 45^\circ = 1 , so we get,
1=ABBC AB=BC \Rightarrow 1 = \dfrac{{AB}}{{BC}}\\\ AB = BC
We know that AB=hAB = h
So, BC=hBC = h

And similarly,
In ΔABD\Delta ABD we have,
tan30=ABBD\tan 30^\circ = \dfrac{{AB}}{{BD}}
And from the figure we know that,
BD=BC+CDBD = BC + CD
We know that BC=hBC = h and CD=40mCD = 40{\rm{ m}} , so we get,
BD=h+40BD = h + 40
Substituting this value of BD in the formula we get,
tan30=hh+40\Rightarrow \tan 30^\circ = \dfrac{h}{{h + 40}}
Since the value of tan30=13\tan 30^\circ = \dfrac{1}{{\sqrt 3 }} , so we get,
13=hh+40\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{h}{{h + 40}}
Solving this equation for the value of hh we get,
3h=h+40 h(31)=40 h=40(1.7321) h=54.64 \Rightarrow \sqrt 3 h = h + 40\\\ \Rightarrow h\left( {\sqrt 3 - 1} \right) = 40\\\ \Rightarrow h = \dfrac{{40}}{{\left( {1.732 - 1} \right)}}\\\ \Rightarrow h = 54.64
Therefore, the height of the building is 54.64m54.64{\rm{ m}} .

Note : It should be noted that when the angle of the elevation is given and we have to find either the height of the building or the distance from the building on the ground we always use the tanθ\tan \theta relationship formula, where θ\theta is the elevation angle.