Question

The shadow of the tower standing on a level ground is x m long when the Sun's altitude is ${30^ \circ }$ , while it is y m long when the altitude is ${60^ \circ }$ . If the height of the tower is $\dfrac{{45\sqrt 3 }}{2}$ m, then x−y is

Answer 1

Hint : Here, we will make a diagram according to the given information, so that we can understand the question more properly. Then by using Pythagoras theorem, we can get the value of tan = Opposite/ Adjacent. Using this tan value, we can find the values of x and y. Finally, we will substitute the values to get the final output of x – y.

Complete step-by-step answer :
According to the given information in the question, we can make the figure as below:

Let the given tower be AD and the height of the tower is given.
$AD = \dfrac{{45\sqrt 3 }}{2}$ m
Given that,
When the angle of elevation is ${30^ \circ }$ , the length of the shadow is x metres.
$\Rightarrow \angle DCA = {30^ \circ }$ and AC = x metres.
Also, given that,
When the angle of elevation is ${60^ \circ }$ , the length of the shadow is y metres.
$\Rightarrow \angle DBA = {60^ \circ }$ and AB = y metres.
We know that, tan C = side opposite to angle C / side adjacent to angle C
Now, we will find the values of x and y to get the final output required.

First we will find the value of x:
In right angle triangle $\Delta DAC$ ,
$\tan C = \dfrac{{AD}}{{AC}}$
Here, the angle of elevation is ${30^ \circ }$ and so substituting the values we will get,
$\tan {30^ \circ } = \dfrac{{\dfrac{{45\sqrt 3 }}{2}}}{x}$
$\Rightarrow \tan {30^ \circ } = \dfrac{{45\sqrt 3 }}{{x \times 2}}$
Since we know that $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$ .
Substituting the value of tan we will get,
$\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{45\sqrt 3 }}{{x \times 2}}$
On evaluating the above equation, we will get,
$\Rightarrow x = \dfrac{{45\sqrt 3 }}{{\dfrac{1}{{\sqrt 3 }} \times 2}}$
$\Rightarrow x = 45\sqrt 3 \div \dfrac{2}{{\sqrt 3 }}$
$\Rightarrow x = 45\sqrt 3 \times \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow x = \dfrac{{45(3)}}{2}$ $(\because \sqrt 3 \times \sqrt 3 = 3)$
$\Rightarrow x = \dfrac{{135}}{2}$

Second, we will find the value of y:
In right angle triangle $\Delta DAB$
$\tan B = \dfrac{{AD}}{{AB}}$
Here, the angle of elevation is ${60^ \circ }$ and so substituting the values we will get,
$\Rightarrow \tan {60^ \circ } = \dfrac{{45\sqrt 3 }}{{y \times 2}}$
We know that $\tan {60^ \circ } = \dfrac{{\dfrac{{45\sqrt 3 }}{2}}}{y}$
Since we know that $\tan {60^ \circ } = \sqrt 3$ .
Substituting the value of tan in the above equation, we will get,
$\Rightarrow \sqrt 3 = \dfrac{{45\sqrt 3 }}{{y \times 2}}$
On evaluating the above equation, we will get,
$\Rightarrow y = \dfrac{{45\sqrt 3 }}{{\sqrt 3 \times 2}}$
$\Rightarrow y = \dfrac{{45}}{2}$
Lastly we need to find the value of x – y:
$\therefore x - y$
Substituting the values of x and y in the above expression, we will get,
$= \dfrac{{135}}{2} - \dfrac{{45}}{2}$
Taking LCM as $2$ , we will get,
$= \dfrac{{135 - 45}}{2}$
Simplify this above expression, and it is reduce in the following form
$= \dfrac{{90}}{2}$
$= 45$ metres
Hence, if the height of the tower is $\dfrac{{45\sqrt 3 }}{2}$ m, then the value of $x - y = 45$ m.
So, the correct answer is “ $x - y = 45$ m.”.

Note : Pythagoras’ Theorem describes the mathematical relationship between three sides of a right-angled triangle. Trigonometry is a field of study in mathematics which observes the relationships of the sides and angles of triangles. The symbol $\theta$ is used to describe an unknown angle. The trigonometric functions sine, cosine and tangent, or as they are more commonly known as sin, cos and tan, can be used to find missing sides or angles of triangles. These functions are defined as the ratios of the different sides of a triangle.