Question

Two buildings are in front of each other on either side of a road of width 10metre. From the top of the first building which is 40 metre high, the angle of elevation to the top of the first building which is 40 metre high, the angle of elevation to the top of the second is $45^\circ $. What is the height of the second building?

Answer 1

Hint:
We can draw a diagram with the given details. Then we can split the height of the building into 2 parts. Then we can identify a right-angled triangle and take the tangent of the angle of elevation and obtain the height of the 2nd building from the distance between the two buildings. Then we can add this height to the height of the smaller building to get the required height of the building.

Complete step by step solution:
We can draw a figure with the details given in the question.

We need to find the height of the building AC. It is given that the building ED is 40 m high.
From the figure we can say that,
$BC = DE = 40m$
$CD = BE = 10m$
We can find from the figure that $AC = AB + BC$
We are given that angle of elevation to the top of the second building from the 1st building is $45^\circ $,
So, from the figure we can write,
$ \Rightarrow \angle AEB = 45^\circ $
Now consider the right-angled triangle ABE right angled at B.
We can take the tan of the angle $\angle AEB$.
We know that $\tan \theta = \dfrac{{{\text{opp}}{\text{. }}\,{\text{side}}}}{{{\text{adj}}{\text{. side}}}}$
From the figure, we can write
$ \Rightarrow \tan \angle AEB = \dfrac{{AB}}{{{\text{AE}}}}$
On substituting the values, we get,
$ \Rightarrow \tan 45^\circ = \dfrac{{AB}}{{10}}$
We know that $\tan 45^\circ = 1$.
$ \Rightarrow 1 = \dfrac{{AB}}{{10}}$
On cross multiplying, we get,
$ \Rightarrow AB = 10$
Now the height of the building is given by,
$AC = AB + BC$
On substituting the values, we get,
\[ \Rightarrow AC = 10 + 40\]
On adding we get,
\[ \Rightarrow AC = 50m\]

So, the required height of the building is 50m.

Note:
Alternate method to solve this problem is,
Now consider the right-angled triangle ABE right angled at B.
It is given that the angle of elevation $\angle AEB = 45^\circ $.
We can use angle sum property in the triangle,
$ \Rightarrow \angle AEB + \angle ABE + \angle BAE = 180^\circ $
On substituting the values, we get,
$ \Rightarrow 45^\circ + 90^\circ + \angle BAE = 180^\circ $
On rearranging we get,
$ \Rightarrow \angle BAE = 180^\circ - 135^\circ $
On simplification we get,
$ \Rightarrow \angle BAE = 45^\circ $
Hence we have,
\[ \Rightarrow \angle AEB = \angle BAE\]
We know that sides opposite to equal angles in a triangle will be equal.
$ \Rightarrow AB = BE$ … (a)
Now the height of the building is given by,
$AC = AB + BC$
On applying equation (a), we get,
$ \Rightarrow AC = BE + BC$
On substituting the values, we get,
\[ \Rightarrow AC = 10 + 40\]
On adding we get,
\[ \Rightarrow AC = 50m\]