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This is a common setup for kinematics problems in physics. My geometry is rusty and I want to understand this very simple idea.
I am having trouble understanding why the angle $\theta$ formed by $\overrightarrow{w}$ is equal to $\theta$ = $\angle$ BOA.
My initial ideas:
My answer is essentially the same as the one given by half-integer fan, but I'll add a picture in case it will aid your understanding. I have labeled new points $C$ and $D$ so that I can refer to them.
$\triangle OCD$ is a right triangle, as is $\triangle OCE$. Because $$\angle OCD=90^\circ=\angle OCE+\angle DCE=\angle OCE+\alpha$$ and $$\angle OCE+\angle COE+90^\circ=\angle OCE+\theta+90^\circ=180^\circ$$ we have that $$\angle OCE +\theta=90^\circ\quad \text{ and }\quad \angle OCE+\alpha=90^\circ,$$ so we must have $\theta=\alpha$.
△BOA (from original image) and △COE (from your image). Am I correct or am I missing something?
$\endgroup$
– jaynp
Jan 2 '13 at 2:45
Since $w \parallel AB$ it follows that
$$\theta + 90^\circ+ \angle B =180^\circ \,.$$
Now, use that $\angle B= 90^\circ- \theta$.
Your approach is correct. If you extend $W$ downward to make a right triangle, the angle opposite to $\angle BOA$ will be $90 - \angle BOA$. Since the angle between the parallel and normal components of $W$ is also $90$, that means $\theta = 90 - (90- \angle BOA) = \angle BOA$.
Not rigorous by any means, but notice that the two angles open at the same rate.
