Answer
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Hint: For solving this problem, first we draw a relatable diagram according to the problem statement. Now the radius of the balloon which subtends an angle α from the eye of an observer and the angle of elevation of its center is β. Consider two triangles $\Delta OAP$ and $\Delta OPC$ After applying the trigonometric ratio of sin in both the triangles we obtain a relationship between all the variables involved. In this way, we can easily prove the equivalence.
Complete step-by-step answer:
By analyzing the statement from the question, the possible figure is:
According to the problem, we are given a balloon of radius r which subtends an angle $\alpha $ from the eye of an observer and the angle of elevation of its center is $\beta $.
Let O be the center of the balloon and P be the eye of the observer. So, the angle subtended by the balloon at the eye is $\angle APB=\alpha $.
$\angle APO=\angle BPO=\dfrac{\alpha }{2}$.
Now considering $\Delta OAP$ and applying trigonometric ratio of sine, we get
$\sin \dfrac{\alpha }{2}=\dfrac{OA}{OP}$
As, OA is the radius of the balloon.
$\begin{align}
& \sin \dfrac{\alpha }{2}=\dfrac{r}{OP} \\
& OP=r\csc \dfrac{\alpha }{2}\ldots (1) \\
\end{align}$
Now, considering the $\Delta OPC$,
$\begin{align}
& \sin \beta =\dfrac{OL}{OP} \\
& OL=OP\sin \beta \ldots (2) \\
\end{align}$
Now, putting the value of OP from equation (1) into equation (2), we get
$OL=r\csc \dfrac{\alpha }{2}\sin \beta $.
So, we proved the equivalence of the expression as the right-hand side is equal to the left-hand side. Hence, we obtain the desired expression as given in the problem statement.
Note: The key step for solving this problem is the diagrammatic representation of the problem statement. Once the student understands the diagram, he can easily apply the trigonometric ratios in the desired triangle to obtain the result.
Complete step-by-step answer:
By analyzing the statement from the question, the possible figure is:
According to the problem, we are given a balloon of radius r which subtends an angle $\alpha $ from the eye of an observer and the angle of elevation of its center is $\beta $.
Let O be the center of the balloon and P be the eye of the observer. So, the angle subtended by the balloon at the eye is $\angle APB=\alpha $.
$\angle APO=\angle BPO=\dfrac{\alpha }{2}$.
Now considering $\Delta OAP$ and applying trigonometric ratio of sine, we get
$\sin \dfrac{\alpha }{2}=\dfrac{OA}{OP}$
As, OA is the radius of the balloon.
$\begin{align}
& \sin \dfrac{\alpha }{2}=\dfrac{r}{OP} \\
& OP=r\csc \dfrac{\alpha }{2}\ldots (1) \\
\end{align}$
Now, considering the $\Delta OPC$,
$\begin{align}
& \sin \beta =\dfrac{OL}{OP} \\
& OL=OP\sin \beta \ldots (2) \\
\end{align}$
Now, putting the value of OP from equation (1) into equation (2), we get
$OL=r\csc \dfrac{\alpha }{2}\sin \beta $.
So, we proved the equivalence of the expression as the right-hand side is equal to the left-hand side. Hence, we obtain the desired expression as given in the problem statement.
Note: The key step for solving this problem is the diagrammatic representation of the problem statement. Once the student understands the diagram, he can easily apply the trigonometric ratios in the desired triangle to obtain the result.
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