From the top of a building 15 m high the angle of elevation of the top of a tower is found to be 30. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60. Find the height of the tower and the distance between the tower and building.
VerifiedHint: Draw the figure according to the information provided in the question and they analyse it to solve the question and assume the height of the Tower be, AD = h, and let the height of the building be CE = 15 m.
Complete step-by-step answer:
According to the question, CE is a building of height 15m, and AD is the tower.
Let us assume the height of the tower be h.
The figure for the above question is shown below-
In triangle ADE,
\[ \tan 60 = \dfrac{{AD}}{{DE}} \]
\[ \therefore \tan 60 = \dfrac{h}{{DE}} \]
\[ \Rightarrow DE = \dfrac{h}{{\sqrt 3 }} \]
From the figure we can see, BCDE is a rectangle.
$ \Rightarrow BC = DE = \dfrac{h}{{\sqrt 3 }}$
And BD = CE.
So, now in triangle ABC-
\[ \tan 30 = \dfrac{{AB}}{{BC}} \]
\[ \Rightarrow \tan 30 = \dfrac{{AB}}{{BC}} = \dfrac{{AB}}{{\dfrac{h}{{\sqrt 3 }}}} \]
\[ \Rightarrow h = 3AB \]
\[ \Rightarrow AB = \dfrac{h}{3} \]
Now, it is given that CE = 15 m.
\[ \Rightarrow AD - AB = 15 \]
\[ \Rightarrow h - \dfrac{h}{3} = 15 \]
\[ \Rightarrow \dfrac{{2h}}{3} = 15 \]
\[ \Rightarrow h = 22.5m \]
And $DE = \dfrac{h}{{\sqrt 3 }} = \dfrac{{22.5}}{{\sqrt 3 }} = 13m$
Therefore, the height of the tower is 22.5 m and the distance between the building and tower is 13 m.
Note: Whenever such types of questions appear, always write the information given in the question, and using that make a the figure, find the value of the tangent of angle ${60^ \circ },{30^ \circ }$, and then solve further by making required substitutions.
