# 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.

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 $\theta=\alpha$.

• Thank you very much. Although it seems that you did not use the same method as half-integer fan. His used the similar triangle △BOA (from original image) and △COE (from your image). Am I correct or am I missing something? – 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. 