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Law of Cosines: Solve for angles given the sides of triangle

Elissa89

Member
Oct 19, 2017
52
I've attached the problem and my work. When I enter cos^1(6.890625) I get an error, but 6.9 is also not the answer and Does Not Exist is also not an acceptable answer. So where I am going wrong with this?

Screen Shot 2018-09-24 at 9.22.02 AM.png

IMG_5340-1.JPG
 
Last edited by a moderator:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
Your image is sideways and hard to read.

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?
 

Elissa89

Member
Oct 19, 2017
52
Your image is sideways and hard to read.

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?
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
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, 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 \(\displaystyle \angle C\).

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

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,713
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.
Hi Elissa89 ,

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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,713
From what I've always seen, triangles are labeled thusly:
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: $a^2=b^2+c^2 -2bc\cos A$.
 

Olinguito

Well-known member
Apr 22, 2018
251
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:

aZnGtMh.jpg
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
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: $a^2=b^2+c^2 -2bc\cos A$.
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. :(
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,713
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:
That's odd.
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 Auto rotate according to EXIF info is checked.
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
Since I caused so much confusion, I wanted to redeem myself by posting a complete solution. Here is a diagram drawn to scale:

mhb_0003.png

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

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

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