A man is carrying a rod which has $2$ baskets at the edge that weights $10N$ and $6N$ like those depicted in the picture. If $AB= 8$ feet then find the distance of $C$ from A where $C$ is such a point that if the man carries the rod with point $C$ on his shoulder then the rod will remain parallel to the horizontal line.
Well that would be $8×6/10$ from the point A. So C will be $3$ feet further from $A$. Thus if the man carries the rod exactly $3$ feet away from point $A$ then it would remain horizontal.
Now I was thinking had not carried the rod exactly at point $C$, rather suppose at point $D$ which is $4$ feet further from points $A$ and $B$. Then the rod would make some angle with the horizontal line.
Now I have tried to find the angle theta here. But failed. Any hint on how to find theta?
Hint: Use the law of sines in right triangles. $$\sin \theta=\frac{\mathbb{opp}}{\mathbb{hyp}}$$
You know that the rod is the hypotenuse.