Question

A portion of a 60 m long tree was broken by a tornado and the top struck up the ground making an angle of \[30{}^\circ \] with the ground level. The height of the point where the tree is broken is equal to
(A) 30 m
(B) 35 m
(C) 40 m
(D) 20 m

Answer 1

Hint: Assume that AB is the length of the tree and it is broken at point C by tornado such that AC is equal to x m. We can write the length of the tree AB as \[AB=AC+CB\] . Now, get the distance CB. The top struck up the ground making an angle of \[30{}^\circ \] with the ground level.
So, \[\angle CBA=30{}^\circ \] . We know the formula, \[\sin \theta =\dfrac{Height}{Hypotenuse}\] . Now, apply sine ratio at \[\angle CBA=30{}^\circ \] . We get the equation \[\dfrac{1}{2}=\dfrac{x}{\left( 60-x \right)}\] . Now, solve this equation further and get the value of x.

Complete step-by-step answer:
First of all, let us assume that AB is the length of the tree and it is broken at point C by tornado such that AC is equal to x m.
According to the question we have the length of the tree AB.
The length of the tree AB = 60 meters ……………………(1)
The length of the tree AB can be written as the summation of the length AC and CB, and the length AC is x m. So,
\[\Rightarrow AB=AC+CB\]
\[\Rightarrow AB=x+CB\] ……………………….(2)
Now, from equation (1) and equation (2), we get
\[\Rightarrow 60=x+CB\]
\[\Rightarrow 60-x=CB\] ……………………………(3)
The top struck up the ground making an angle of \[30{}^\circ \] with the ground level.
So, \[\angle CBA=30{}^\circ \] .

In \[\Delta CAB\] , we have,
\[CA=x\]
\[CB=\left( 60-x \right)\]
\[\angle CBA=30{}^\circ \]
We know the formula, \[\sin \theta =\dfrac{Height}{Hypotenuse}\] .
Now, applying sine ratio at angle \[30{}^\circ \] in \[\Delta CAB\] ,
\[\Rightarrow \sin 30{}^\circ =\dfrac{AC}{CB}\]
\[\Rightarrow \sin 30{}^\circ =\dfrac{x}{\left( 60-x \right)}\] ………………….(9)
We know that \[\sin 30{}^\circ =\dfrac{1}{2}\] ………………………….(10)
Now, from equation (9) and equation (10), we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}=\dfrac{x}{\left( 60-x \right)} \\
 & \Rightarrow \left( 60-x \right)=2x \\
 & \Rightarrow 60=x+2x \\
 & \Rightarrow 60=3x \\
 & \Rightarrow \dfrac{60}{3}=x \\
 & \Rightarrow 20=x \\
\end{align}\]
The distance \[AC=x=20m\] .
Therefore, the length at which the tree is broken from the ground is 20 m.
Hence, the correct option is (D).

Note: In this question, one can think to use cosine ratio at the angle \[30{}^\circ \] .

Now, applying cosine ratio at angle \[30{}^\circ \] in \[\Delta CAB\] ,
\[\Rightarrow \cos 30{}^\circ =\dfrac{AB}{CB}\]
\[\Rightarrow \cos 30{}^\circ =\dfrac{AB}{\left( 60-x \right)}\]
Here, we don’t have any idea about the length AB is unknown. So, we cannot apply the cosine ratio here. Therefore, we should apply sine ratio here.