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:
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:
- Why did my book assume that the constants of proportionality were equal? Is that included in the definition of the word 'proportional'?
Best Answer
Since the magnitudes and the cosines are in direct proportion, each corresponding-magnitude-cosine pair is in the same ratio. That is, $$\frac P{\cos A}=\frac Q{\cos B}.$$
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.