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'?