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.
5 Answers
$\begingroup$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 $\theta=\alpha$.
$\endgroup$ 1 $\begingroup$Since $w \parallel AB$ it follows that
$$\theta + 90^\circ+ \angle B =180^\circ \,.$$
Now, use that $\angle B= 90^\circ- \theta$.
$\endgroup$ $\begingroup$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 $\theta = 90 - (90- \angle BOA) = \angle BOA$.
$\endgroup$ $\begingroup$Not rigorous by any means, but notice that the two angles open at the same rate.
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
$\endgroup$ 1
