Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a common setup for kinematics problems in physics. My geometry is rusty and I want to understand this very simple idea.

enter image description here

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.
share|improve this question

4 Answers 4

up vote 1 down vote accepted

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.

enter image description here

$\triangle OCD$ 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 $\theta=\alpha$.

share|improve this answer
    
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? –  jp24 Jan 2 '13 at 2:45

Not rigorous by any means, but notice that the two angles open at the same rate.

triangle animation

share|improve this answer

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 $\theta = 90 - (90- \angle BOA) = \angle BOA$.

share|improve this answer

Since $w \parallel AB$ it follows that

$$\theta + 90^\circ+ \angle B =180^\circ \,.$$

Now, use that $\angle B= 90^\circ- \theta$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.