Question

A pole of height $20ft$ has a shadow of length $11.55ft$ at a particular instant of time. Find the angle of elevation (in degrees) of the sun at this point of time?
A) ${90^0}$
B) ${60^0}$
C) ${30^0}$
D) ${45^0}$

Answer 1

Hint:
Draw the diagram according to the description given. Know where exactly is the elevation of the sun. apply $\tan \theta = \dfrac{{{\text{Opposite}}}}{{Adjacent}}$ and find the value of $\theta $ required. Where $\theta $ is the elevation of the sun.

Complete step by step solution:
So we have the height of the pole given as $20ft$ high. And the length of the shadow will be as $11.55ft$ .
So for solving it, we will first draw the figure, in this figure the side $AB$ will be the height and the $BC$ will be the path difference between them.

Let us assume that the $\theta $ be the elevation of the sun, so from this by using the formula of it, we get
$ \Rightarrow \tan \theta = \dfrac{{AB}}{{BC}}$
Now on substituting the values, we get
$ \Rightarrow \tan \theta = \dfrac{{20ft}}{{11.55ft}}$
On changing the unit, we get
$ \Rightarrow \tan \theta = \dfrac{{2000}}{{1155}}$
So we will make the above fraction into the simplest form, therefore, we get
$ \Rightarrow \tan \theta = \dfrac{{400}}{{231}}$
Now on solving the above equation, we get
$ \Rightarrow \tan \theta = 1.73$
And it can be written as
$ \Rightarrow \tan \theta = \sqrt 3 $
So from here, the value of $\theta $ can be calculated by taking the $\tan $ to the right side, so we get
$ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right)$
And it can also be written in degree from as ${60^ \circ }$

Therefore, the option $\left( B \right)$ is correct.

Note:
when asked about the angle of elevation of the sun, the angle is opposite to the vertical heights and when asked about the angle of depression of the sun, the angle is opposite to the distance of the path.