Is it correct to use $\vec F=m\vec a\cos\theta$? 
Suppose a force $\vec F$ is applied on an object of mass $m$, which experiences an acceleration $\vec a$. The angle between $\vec F$ and $\vec a$ is $\theta$.
To find the component of $\vec a$ along the direction of $\vec F$, will it be correct to use $\vec F=m\vec a\cos\theta$? If yes, shouldn't $\vec F=m\vec a\cos\theta$ be the general relation between $\vec F$ and $\vec a$ instead of $\vec F=m \vec a$?
 A: If $\vec F$ is the only force acting on the object then $\vec a$ must be parallel to $\vec F$, so $\theta =0$, $\cos \theta=1$, and $\vec F=m\vec a$.
On the other hand, if $\theta$ is not zero then there must be one or more other forces acting on the object so that the net force $\vec F_{net}$ is parallel to $\vec a$, and $\vec F_{net}=m \vec a$. In this case we cannot write down any relationship between $\vec F$ and $\vec a$ unless we know something about the other forces.
A: What you are looking for is the magnitude, the value, of the force in a particular direction. So skip the vector arrow symbols.
You are completely right that $ma\cos(\theta)$ will give you the correct horisontal component you are looking for. In this particular case, the original force is already pointing in the direction of this component, though, so $\cos(\theta)=1$. It is thus not necessary to include this term. Just calculate $ma$ and you have the force magnitude you want.
In fact, how did you draw the right-angled triangle? Typically we would draw it so that the force points along with the hypotenuse. In your case, this would turn out to be no triangle since the vertical leg is zero and the horisontal leg equals the force itself.
A: Newton's second law states that the direction of acceleration is also the same as the direction of the net force. The term $macos(\theta)$ is therefore not to be considered the direction of acceleration due to the force $F$. So $a$ is the correct acceleration.
A: No, that's not correct. $\vec a$ is not directed horizontally, the prefactor $\cos\theta$ doesn't change this fact as it only can change the magnitude, not the direction.
A: You can think this way: for any $\vec{a}$ you can decompose it
$$\vec{a}= a_1\vec{e_1} + a_2\vec{e_2} + a_3\vec{e_3}$$
When you writes $\vec{F}=m \vec{a}$, you're actually saying that
$$ F_i\; \vec{e_i}=ma^i\; \vec{e_i}$$
Basically you need to know what direction matters for you problem and then project $\vec{a}$ in this direction. Thats why $\cos(\theta)$ appear in your problem (but not in the general formulation).
