Question

A pole 6m high casts a shadow of $2\sqrt 3 m$ long on the ground, then the sun’s elevation is?

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 $6m$ high. And the length of the shadow will be as $2\sqrt 3 m$ .
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 it.

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{6}{{2\sqrt 3 }}$
Now on solving the above equation, we get
$ \Rightarrow \tan \theta = \sqrt 3 $
As we know the value of the $\theta $ when it is $\sqrt 3 $ , so we get
$ \Rightarrow \tan \theta = \tan \left( {\dfrac{\pi }{3}} \right)$
On equating this, we get
$ \Rightarrow \theta = \dfrac{\pi }{3}$
And it can also be written in degrees from as ${60^ \circ }$ .

So the ${60^ \circ }$ will be the angle of elevation of sun.

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. So by using this we can solve these types of questions.