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Question:

Two concurrent forces act along the sides CA and CB of a triangle. Their magnitudes are proportional to $\cos (A)$ and $\cos (B)$ respectively. Prove that their resultant is proportional to $\sin(C)$.

My book's attempt:

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Let the forces acting along $CA$ and $CB$ be $\vec{P}$ and $\vec{Q}$ respectively. Let the resultant of $\vec{P}$ and $\vec{Q}$ be $\vec{R}$, which is acting along $CD$. Now, let $P=k\cos(A)$, $Q=k\cos(B)$, where $k$ is the constant of proportionality.

$$....$$

$$[\text{Then my book went on to prove that}\ R=k\sin(C)]$$

My comments:

If $P$ is proportional to $\cos(A)$, it means that $P=n\cos(A)$. Again, if $Q$ is proportional to $\cos(B)$, it means that $Q=s\cos(B)$. Furthermore, if $R$ is proportional to $\sin(C)$, it means that $R=t\sin(C)$. Now, the constants of proportionality $n,s,t$ needn't be equal. Why did my book assume that the constants were equal i.e. why did my book assume:

$$n=s=t=k$$

My question:

  1. Why did my book assume that the constants of proportionality were equal? Is that included in the definition of the word 'proportional'?
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    When it says CA and CB are proportional means that $\frac{CA}{CB}=\frac {\cos A}{\ cos B}=\frac{k \cos A}{k\cos B}$ where as it says k is constant of proportionality. – sirous Feb 28 '22 at 04:22

1 Answers1

1

Question:

Two concurrent forces act along the sides CA and CB of a triangle. Their magnitudes are proportional to $\cos (A)$ and $\cos (B)$ respectively.

My book's attempt:

Let the forces acting along $CA$ and $CB$ be $\vec{P}$ and $\vec{Q}$ respectively. Let the resultant of $\vec{P}$ and $\vec{Q}$ be $\vec{R}.$

Since the magnitudes and the cosines are in direct proportion, each corresponding-magnitude-cosine pair is in the same ratio. That is, $$\frac P{\cos A}=\frac Q{\cos B}.$$

My question:

If $P$ is proportional to $\cos(A)$, it means that $P=n\cos(A)$. Again, if $Q$ is proportional to $\cos(B)$, it means that $Q=s\cos(B)$. Why did my book assume that the constants of proportionality were equal?

Let $m_1,m_2,m_3$ represent the three magnitudes and $c_1,c_2,c_3$ correspondingly represent the three cosines. I.e., $m_1=P, c_1=\cos A,$ etc.

To be clear, the given information is $$m\propto c,\\\text{i.e., }\frac{m_1}{c_1}=k=\frac{m_2}{c_2},$$ rather than $$m_1\propto c_1,\;m_2\propto c_2,\\\text{i.e., }m_1=k_1c_1,\;m_2=k_2c_2.$$

Analogously, if $x=7y,$ then we say that $x$ is proportional to $y;$ we don't say that $21$ is proportional to $3.$

$m$ and $x$ are variables, while $P$ and $21$ are particular values or constants.

ryang
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  • Would you agree that the question has been formulated rather erroneously? They said, "Their magnitudes are proportional to cos(A) and cos(B) respectively.", which means that P is proportional to cos(A) and Q is proportional to cos(B). Shouldn't they have said that "the magnitudes and the cosines are in direct proportion", as you did? Furthermore, in the end, it said, "Prove that their resultant is proportional to sin(C).", which is like saying, "prove that 21 is proportional to 3". Do you concur? – tryingtobeastoic Mar 01 '22 at 05:36
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    Yes I agree. A more familiar example: Triangles A and B are simliar to each other iff the length of A's sides is proportional to the length of B's sides iff the length of A's sides and the length of B's sides are in proportion. Also see Ratio vs. Proportion. – ryang Mar 01 '22 at 07:14
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    To be fair though, the given problem statement really has only one meaningful interpretation. As for its question, I guess we might carefully rephrase it as “prove that the resultant force and $\sin C$ are governed by the same proportionality”. – ryang Mar 01 '22 at 07:45
  • Yeah, that was on my mind as well. The blatant error was when they said, "Their magnitudes are proportional to cos(A) and cos(B) respectively." The last sentence, "Prove that their resultant is proportional to sin(C)." is correct I think (I'm retracting what I said earlier about the last sentence of the question). What do you think? (Maybe you have a typo in your 2nd latest comment) – tryingtobeastoic Mar 01 '22 at 08:22
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    No typo. The question sentence is (technically) not meaningfully written. – ryang Mar 01 '22 at 08:50
  • "As for its question, I guess we might carefully rephrase it as “prove that the resultant force and sinC are governed by the same proportionality”"- are you saying this because the resultant force and sinC are both constants (in the same way that 21 and 3 are both constants and x and y are the variables, and we can say that x is proportional to y, but can't say 21 is proportional to 3; we have to say that 21 and 3 are governed by the same proportionality as x and y)? Is the book's question sentence wrong because they are both constants? – tryingtobeastoic Mar 01 '22 at 12:47
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    "$R$ and $\sin C$ are governed by the same proportionality" = "$R$ and $\sin C$ satisfy the same relationship $m=kc.$" – ryang Mar 01 '22 at 12:52