Why are these angles equal for object on inclined plane? 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:


*

*If we extend $w$ we can get a right triangle and somehow prove the angles equal by similarity.

*Some sort of use of interior angles and parallel lines.

 A: 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 $\theta=\alpha$.
A: Since $w \parallel AB$ it follows that
$$\theta + 90^\circ+ \angle B =180^\circ \,.$$
Now, use that $\angle B= 90^\circ- \theta$.
A: 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 $\theta = 90 - (90- \angle BOA) = \angle BOA$.
A: Not rigorous by any means, but notice that the two angles open at the same rate.

A: As @Zev Chonels said
∠OCD = 90∘ = ∠OCE + α
therefore ∠OCE = 90 - α
and
∠OCE + θ + 90∘ = 180∘    (sums of all angle of triangle = 180∘)
substitute the value of ∠OCE
90∘ - α + θ + 90∘ = 180∘
θ - α + 180∘ = 180∘
θ - α = 180∘ - 180∘
θ - α = 0
θ = α
