
If ${{T}_{1}}=\dfrac{mg}{\sin \alpha +\dfrac{\cos \alpha }{\cos \beta }\sin \beta }$
What is the final answer?
Answer
558.9k+ views
Hint: There is not much information given in the question to get the solution, so don’t assume anything. Just solve the question by considering it a mathematical equation. One thing that can be deduced from the equation is that it might be talking about tension of a string as it is related to mass and gravity.
Formula used:
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Complete step by step solution:
Given that,
${{T}_{1}}=\dfrac{mg}{\sin \alpha +\dfrac{\cos \alpha }{\cos \beta }\sin \beta }$
As we know that T represents the tension; it is a force along the length of a medium, especially a force carried by a flexible medium as the rope.
Now, to solve the equation take L.C.M of the terms in the denominator,
So, after taking the L.C.M we can rewrite equation as:
$\begin{align}
& {{T}_{1}}=\dfrac{mg}{\dfrac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \beta }} \\
& \therefore {{T}_{1}}=\dfrac{mg\cos \beta }{\sin \alpha \cos \beta +\cos \alpha \sin \beta } \\
\end{align}$
We know the trigonometric identity states that $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Using this identity, we can rewrite the above equation as follows:
${{T}_{1}}=\dfrac{mg\cos \beta }{\sin \left( \alpha +\beta \right)}$
Therefore, after solving the equation given in the question, we get –
${{T}_{1}}=\dfrac{mg\cos \beta }{\sin \left( \alpha +\beta \right)}$, this is the final answer as there is not much information given in the question.
Note:
This type of question is very rare in papers, as there is not information given for the solution in the question, so it’s better we don’t assume anything. To get the solution for this question, it's better we just treat it as a mathematical equation that needs to be solved and reduced to a simpler form. The general mistake that students do is, they assume the equation as to find the tension of the string and then assume some value to get the answer, but that approach is wrong, as nothing should be assumed until something is explicitly mentioned in the question.
Formula used:
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Complete step by step solution:
Given that,
${{T}_{1}}=\dfrac{mg}{\sin \alpha +\dfrac{\cos \alpha }{\cos \beta }\sin \beta }$
As we know that T represents the tension; it is a force along the length of a medium, especially a force carried by a flexible medium as the rope.
Now, to solve the equation take L.C.M of the terms in the denominator,
So, after taking the L.C.M we can rewrite equation as:
$\begin{align}
& {{T}_{1}}=\dfrac{mg}{\dfrac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \beta }} \\
& \therefore {{T}_{1}}=\dfrac{mg\cos \beta }{\sin \alpha \cos \beta +\cos \alpha \sin \beta } \\
\end{align}$
We know the trigonometric identity states that $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Using this identity, we can rewrite the above equation as follows:
${{T}_{1}}=\dfrac{mg\cos \beta }{\sin \left( \alpha +\beta \right)}$
Therefore, after solving the equation given in the question, we get –
${{T}_{1}}=\dfrac{mg\cos \beta }{\sin \left( \alpha +\beta \right)}$, this is the final answer as there is not much information given in the question.
Note:
This type of question is very rare in papers, as there is not information given for the solution in the question, so it’s better we don’t assume anything. To get the solution for this question, it's better we just treat it as a mathematical equation that needs to be solved and reduced to a simpler form. The general mistake that students do is, they assume the equation as to find the tension of the string and then assume some value to get the answer, but that approach is wrong, as nothing should be assumed until something is explicitly mentioned in the question.
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