Question

A boy saw the top of a building under construction at an elevation of $ {30^ \circ } $ . The completed building was 12m higher and the boy saw its top at an elevation of $ {60^ \circ } $ from the same spot.
A. Draw a rough figure based on the given details.
B. What is the height of the building?
C. What is the distance between the building and the boy?

Answer 1

Hint:This problem deals with the applications of trigonometry. The applications of trigonometry are used to define the relationship between the sides of the right angled triangle considered in the given problem. Here in a considered right angled triangle we can obtain the relation between the sides of the triangle with the help of trigonometric ratios. Here in this problem these ratios are used:
 $ \Rightarrow \tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }} $ , and
 $ \Rightarrow \tan {60^ \circ } = \sqrt 3 $

Complete step-by-step answer:
Given that a boy is standing at a point, let’s say B. So the boy is standing at the point B and saw a building which is under construction at an elevated angle $ {30^ \circ } $ .
Let us consider that the height of the building OP, which is under construction be $ x $ .
Here the height of the incomplete building is OP = $ x $ .
Now when the construction of the building is completed and the boy who is standing at the same point B sees that the building is 12 m higher than before completion of the building, now the elevation angle is $ {60^ \circ } $ .
Let us consider that after the construction of the building the building height is increased by 12m, given by PA = 12 m.
So here the height of the complete building is given by:
 $ \Rightarrow $ OP + PA = $ x + 12 $
 $ \therefore $ The height of the complete building is $ x + 12 $ .
A rough figure based on the given details is given below:

What is the height of the building?
Applying trigonometric ratio to the angle $ {30^ \circ } $ , as given below:
 $ \Rightarrow \tan {30^ \circ } = \dfrac{{OP}}{{OB}} $
 $ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{x}{{OB}} $
 $ \therefore OB = x\sqrt 3 $
Now applying trigonometric ratio to the angle $ {60^ \circ } $ , as given below:
 $ \Rightarrow \tan {60^ \circ } = \dfrac{{OA}}{{OB}} $
 $ \Rightarrow \sqrt 3 = \dfrac{{x + 12}}{{OB}} $
 $ \therefore OB = \dfrac{{x + 12}}{{\sqrt 3 }} $
Now equating both the expressions of $ OB $ , as given below:
Equating $ OB = x\sqrt 3 $ and $ OB = \dfrac{{x + 12}}{{\sqrt 3 }} $
 $ \Rightarrow x\sqrt 3 = \dfrac{{x + 12}}{{\sqrt 3 }} $
 $ \Rightarrow \sqrt 3 \left( {x\sqrt 3 } \right) = x + 12 $
Simplifying the equation, we know that $ \sqrt 3 \times \sqrt 3 = 3 $ .
 $ \Rightarrow 3x = x + 12 $
 $ \Rightarrow 2x = 12 $
 $ \Rightarrow x = 6 $
So the total height of the building is given by $ x + 12 $ , as given below:
 $ \Rightarrow x + 12 = 6 + 12 $
 $ \Rightarrow 18 $
 $ \therefore $ The height of the building is 18 m.
What is the distance between the building and the boy?
The distance between the boy and the building is the distance between the points O and B, which is the length of OB.
From the above obtained expression, $ OB = x\sqrt 3 $ , substitute the value of $ x $ , to get the value of OB, as given below:
 $ \Rightarrow OB = 6\left( {\sqrt 3 } \right) $
 $ \therefore OB = 6\sqrt 3 $ m.

Hence the distance between the building and the boy is $ 6\sqrt 3 $ m.

Note:
Please note that this problem can also be done in another way but with a slight change. That is instead of equating the expressions of OB to get the value of $ x $ . We could just divide the obtained expressions for $ \tan {30^ \circ } $ and $ \tan {60^ \circ } $ , and then simplify the equation, here anyway OB gets cancelled. Hence obtaining the same end result.