Question

If a flag staff of 6m height is placed on the top of the tower throws a shadow of $2\sqrt{3}$metres along the ground then the angle that the sun makes with the ground is:
(a) ${{60}^{\circ }}$
(b) ${{30}^{\circ }}$
(c) ${{45}^{\circ }}$
(d) None of these

Answer 1

Hint:First, before proceeding for this, we must draw the above in form of a triangle where the sun is at point A and making the shadow of the tower with angle supposed as $\theta $. Then, we are supposed to find the value of angle $\theta $made the shadow of the tower on the ground by reflection from the sun at A. Then, by using the concept of trigonometry in the triangle CDE where p is perpendicular and b is base as $\tan \theta =\dfrac{p}{b}$, we get the required value of the angle.

Complete step by step answer:
In this question, we are supposed to find the value of the angle made by the sun when a flagstaff of 6m height is placed on the top of the tower throws a shadow of $2\sqrt{3}$metres.
So, before proceeding for this, we must draw the above in form of a triangle where the sun is at point A and making the shadow of the tower with angle supposed as $\theta $, we get:

Now, we can see clearly that tower is CD in the figure with height 6m and the length of its shadow is $2\sqrt{3}$metres which is DE.
Then, we are supposed to find the value of angle $\theta $made the shadow of the tower on the ground by reflection from the sun at A.
So, by using the concept of trigonometry in the triangle CDE where p is perpendicular and b is base as:
$\tan \theta =\dfrac{p}{b}$
Now, by substituting the value of the p as 6 and b as $2\sqrt{3}$from the figure, we get:
$\tan \theta =\dfrac{6}{2\sqrt{3}}$
Then, by solving the above expression, we get the value as:
$\begin{align}
  & \tan \theta =\sqrt{3} \\
 & \Rightarrow \theta ={{\tan }^{-1}}\sqrt{3} \\
 & \Rightarrow \theta ={{60}^{\circ }} \\
\end{align}$
So, we get the value of angle as ${{60}^{\circ }}$.
Hence, option (a) is correct.

Note:
Now, to solve these type of the questions we need to know some of the basics of the right angled triangle with sides as perpendicular p, base b and hypotenuse h and also by using the concept of trigonometry, we get the formulas for the following figure as:

$\begin{align}
  & \sin \theta =\dfrac{p}{h} \\
 & \cos \theta =\dfrac{b}{h} \\
 & \tan \theta =\dfrac{p}{b} \\
\end{align}$