Conversation started Feb 11, 2023 at 23:34.
D.C. the III
D.C. the III
Feb 11, 2023 23:34
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Ted Shifrin
Ted Shifrin
Right, although I would leave the third side of the triangle in the picture. So what is $1/2$ base times height?
D.C. the III
D.C. the III
So I'm not going to be all high and mighty, but I knew this is what you were intimating at, I didn't just jump to it immediately and what is still troubling me now is that I can't see how I would write $a$ in terms of $\theta$. I see how how the height will be written in terms of $\theta$, but not my base
Ted Shifrin
Ted Shifrin
This formula has shown up in various exercises before in the book, by the way. There's one for sure in the early sections of Chapter 3.
The base is $a$. What's the height?
D.C. the III
D.C. the III
the third side of the triangle meaning the one that "splits" my original quadrilateral? Height will be: $b \sin(\theta - \frac{\pi}{2})$
Ted Shifrin
Ted Shifrin
No. $\sin(\pi-\theta) = \sin\theta$.
Right. So we can see the triangle whose area we are computing :)
The formula (which you learned in trig and probably haven't used enough) is $\frac12 ab\sin\theta$.
D.C. the III
D.C. the III
Feb 11, 2023 23:41
oh shoot your right. $\theta - \frac{\pi}{2}$ is the angle for the third side
@TedShifrin Lol...if ever.....I recognize it, but I don't think I ever used it consiously as the area formula
Ted Shifrin
Ted Shifrin
This very formula was needed to do #11 in the same section. If I didn't assign that, I probably did it in lecture.
D.C. the III
D.C. the III
yea not assigned
Ted Shifrin
Ted Shifrin
It comes in also with the area of the parallelogram that appears in the magnitude of the cross-product, although I purposely did not do that in the book.
But if you ask most any calc III or physics student, they'll tell you $\|\vec a\times\vec b\| = \|\vec a\| \|\vec b\|\sin\theta$.
D.C. the III
D.C. the III
It clicked why I don't need to express $a$ in terms of $\theta$, it is already a fixed constant. It was only height that needed recharacterizing.
Ted Shifrin
Ted Shifrin
Yuppers. Of course, if you decide to use $b$ as the base, then the height is $a\sin\theta$. No problem.
D.C. the III
D.C. the III
Feb 11, 2023 23:47
I enjoy your geometry questions even though this probably should not have taken as long as it did. My faulty (albeit gradually improving) geometry is what is hindering me
 
Conversation ended Feb 11, 2023 at 23:47.