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

In summary: The angles in the diagram are labeled using standard notation.To find \angle A, we would then use:A=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right)\approx34.5^{\circ}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^
  • #1
Elissa89
52
0
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?

View attachment 8436

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  • #2
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?
 
  • #3
MarkFL said:
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.
 
  • #4
Elissa89 said:
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:

\(\displaystyle A=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right)\approx34.5^{\circ}\)
 
  • #5
Elissa89 said:
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.
 
  • #6
I like Serena said:
...How is that not the standard practice MarkFL?...

From what I've always seen, triangles are labeled thusly:

labelled-triangle-diagram.png
 
  • #7
MarkFL said:
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: $a^2=b^2+c^2 -2bc\cos A$.
 
  • #8
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:

 

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  • #9
I like Serena said:
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. :(
 
  • #10
Olinguito said:
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.
 
  • #11
Since I caused so much confusion, I wanted to redeem myself by posting a complete solution. Here is a diagram drawn to scale:

View attachment 8441

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

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

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

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1. What is the Law of Cosines?

The Law of Cosines is a mathematical formula used to determine the relationship between the sides and angles of a triangle. It is used to solve for missing angles or sides in a triangle when given enough information.

2. How is the Law of Cosines different from the Pythagorean Theorem?

The Pythagorean Theorem is used to solve for the length of one side of a right triangle when given the other two sides. The Law of Cosines, however, can be used to solve for any missing side or angle of a triangle, regardless of whether it is a right triangle or not.

3. When should the Law of Cosines be used?

The Law of Cosines is typically used when given three sides of a triangle and trying to solve for one of the angles. It can also be used when given two sides and the included angle and trying to solve for the third side. However, it cannot be used when given only two angles and one side.

4. What is the formula for the Law of Cosines?

The formula for the Law of Cosines is c² = a² + b² - 2abcosC, where c is the length of the side opposite angle C, and a and b are the lengths of the other two sides of the triangle.

5. Are there any limitations to using the Law of Cosines?

Yes, the Law of Cosines can only be used in triangles, and all three sides must be known or given in order to solve for an angle. It cannot be used when given only two angles and one side, or when the triangle is not a valid triangle (e.g. one side is longer than the sum of the other two sides).

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