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Question: The shadow of a tower is found to be \( 60m \) shorter when the Sun’s altitude changes from \( {30^ ...

The shadow of a tower is found to be 60m60m shorter when the Sun’s altitude changes from 30{30^ \circ } to 60{60^ \circ } . The height of the tower from the ground is approximately:
(A) 62m62m
(B) 301m301m
(C) 101m101m
(D) 75m75m
(E) 52m52m

Explanation

Solution

Hint : The given question belongs to the height and distances concept of trigonometry domain. In the problem, we have to find the height of a tower given that the length of the shadow of the tower is 60m60m shorter when the Sun's altitude changes from 30{30^ \circ } to 60{60^ \circ } . We first draw a diagram for understanding the situation better and then use the concepts of trigonometry to find the height of the tower.

Complete step-by-step answer :

Let the tower be AB and length be y meters.
Now, we are given that the shadow of a tower is 60m60m shorter when the Sun’s altitude changes from 30{30^ \circ } to 60{60^ \circ } .
So, BC is the original shadow of the tower when the altitude of the sun is 30{30^ \circ } and BD is the shadow of the tower when the sun's altitude is 60{60^ \circ } .
So, we get the difference between the lengths of shadow as CD=60mCD = 60m .
Let us assume that BD be of length x meters.
Now, using trigonometry in the triangle ADB,
tan60=PerpendicularBase=ABBD\tan {60^ \circ } = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{AB}}{{BD}}
Substituting the values of AB and BD, we get,
3=yx\Rightarrow \sqrt 3 = \dfrac{y}{x}
y=3x(1)\Rightarrow y = \sqrt 3 x - - - - \left( 1 \right)
Now, in triangle ABC, we have,
tan30=PerpendicularBase=ABBC\tan {30^ \circ } = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{AB}}{{BC}}
Substituting the lengths of Ab and BC, we get,
13=yx+60\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{y}{{x + 60}}
Substituting the value of y from equation (1)\left( 1 \right) in the above expression, we get,
13=3xx+60\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{\sqrt 3 x}}{{x + 60}}
Cross multiplying the terms, we get,
x+60=3x\Rightarrow x + 60 = 3x
Shifting the terms in the equation, we get,
2x=60\Rightarrow 2x = 60
Dividing both the sides of the equation by 22 , we get,
x=30\Rightarrow x = 30
So, we get the value of x as 3030 . Substituting the value of x in equation (1)\left( 1 \right) , we get value of y as,
y=303y = 30\sqrt 3
Hence, the height of the tower is 30330\sqrt 3 meters.
Finding the approximate value, we get 303=51.9630\sqrt 3 = 51.96 .
So, the approximate height of the tower is 5252 meters.
Hence, option (E) is the correct answer.
So, the correct answer is “Option E”.

Note : We should have a strong grip over the concepts of trigonometry and height and distances in order to deal with such kinds of problems. We should have accuracy in calculations, derivatives and arithmetic in order to be sure of the final answer.