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Okay, if I am trying to find \(\displaystyle \angle A\), I would begin with the fact that this angle is opposite side \(c\) and write:

\(\displaystyle c^2=a^2+b^2-2ab\cos(A)\)

\(\displaystyle \cos(A)=\frac{a^2+b^2-c^2}{2ab}\)

\(\displaystyle A=\arccos\left(\frac{a^2+b^2-c^2}{2ab}\right)\approx51.4^{\circ}\)

Can you proceed in like manner to find the other two angles?

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Answer wasn't correct for angle A but it was correct for angle c. Sorry it was sideways, that was just how it uploaded. Still not sure why I can't find angle A though when a=21 is opposite of angle A.

Okay, if I am trying to find \(\displaystyle \angle A\), I would begin with the fact that this angle is opposite side \(c\) and write:

\(\displaystyle c^2=a^2+b^2-2ab\cos(A)\)

\(\displaystyle \cos(A)=\frac{a^2+b^2-c^2}{2ab}\)

\(\displaystyle A=\arccos\left(\frac{a^2+b^2-c^2}{2ab}\right)\approx51.4^{\circ}\)

Can you proceed in like manner to find the other two angles?

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Okay, that's not standard practice, the way the angles and sides are labeled. But, given that it is labeled that way, then what I posted would indeed work for \(\displaystyle \angle C\).Answer wasn't correct for angle A but it was correct for angle c. Sorry it was sideways, that was just how it uploaded. Still not sure why I can't find angle A though when a=21 is opposite of angle A.

To find \(\displaystyle \angle A\), we would then use:

\(\displaystyle A=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right)\approx34.5^{\circ}\)

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- Mar 5, 2012

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Hi Elissa89 ,Answer wasn't correct for angle A but it was correct for angle c. Sorry it was sideways, that was just how it uploaded. Still not sure why I can't find angle A though when a=21 is opposite of angle A.

Your initial formula was correct.

How is that not the standard practice MarkFL ?

You have:

$$a^2=b^2+c^2-2bc\cos A\\

21^2=29^2+37^2-2(29)(37)\cos A \\

441=841+1369-2146\cos A $$

How did you get from there to:

$$\frac{441}{64}=\frac{(\cancel{64})\cos A}{\cancel{64}}$$

Because that is not correct.

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From what I've always seen, triangles are labeled thusly:...How is that not the standard practice MarkFL ?...

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- #7

- Mar 5, 2012

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That's indeed the correct triangle, although I'm used to having A at bottom left.From what I've always seen, triangles are labeled thusly:

Either way, $a$ is opposite the angle $A$ as it should be.

And then the cosine rule is: $a^2=b^2+c^2 -2bc\cos A$.

- Apr 22, 2018

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Okay, now that I cock my head to the side and squint my eyes, I see the triangle is labeled that way. Sorry for all the confusion.That's indeed the correct triangle, although I'm used to having A at bottom left.

Either way, $a$ is opposite the angle $A$ as it should be.

And then the cosine rule is: $a^2=b^2+c^2 -2bc\cos A$.

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- #10

- Mar 5, 2012

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That's odd.The image itself isn’t sideways – I downloaded it and it was the right way up – for some reason it was rotated when attached in the post. I’ve uploaded it to Imgur and retrieved it from there instead:

After downloading I see a thumbnail that is sideways.

But when I open it in an image viewer it is indeed the right way up.

Never seen that before!

Anyway, I can see in my image viewer that the option

And indeed, the EXIF info shows swapped dimensions.

Elissa89 , if I may ask, how did you scale down the image?

I can see that the original size was much bigger.

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- #11

\(\displaystyle A=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right)\approx34.5^{\circ}\)

\(\displaystyle B=\arccos\left(\frac{a^2+c^2-b^2}{2ac}\right)\approx94.1^{\circ}\)

\(\displaystyle C=\arccos\left(\frac{a^2+b^2-c^2}{2ab}\right)\approx51.4^{\circ}\)