# Area of triangle...check work please

• Dec 12th 2007, 08:19 AM
aikenfan
I have been having a hard time with these...could someone please look over my work...I was very confused on how to solve number 4, I am still attempting it as of right now....maybe someone could help me with that one....
http://i103.photobucket.com/albums/m...n91919/1-4.jpg
http://i103.photobucket.com/albums/m...1919/88003.jpg
• Dec 12th 2007, 09:03 AM
topsquark
Quote:

Originally Posted by aikenfan
I have been having a hard time with these...could someone please look over my work...I was very confused on how to solve number 4, I am still attempting it as of right now....maybe someone could help me with that one....
http://i103.photobucket.com/albums/m...n91919/1-4.jpg
http://i103.photobucket.com/albums/m...1919/88003.jpg

I haven't actually checked the work, but it looks like you have applied the equations correctly.

For the fourth one, use the Law of Cosines:
$a^2 = b^2 + c^2 - 2bc \cdot cos(A)$

The other two forms are
$b^2 = a^2 + c^2 - 2ac \cdot cos(B)$

$c^2 = a^2 + b^2 - 2ab \cdot cos(C)$

As you might have guessed, these are generalizations of the Pythagorean Theorem to triangles with arbitrary angles (not just right angles.)

-Dan
• Dec 12th 2007, 03:09 PM
aikenfan
How do I use that formula if I do not have A? I am a little bit confused...
• Dec 13th 2007, 07:06 AM
topsquark
Quote:

Originally Posted by aikenfan
How do I use that formula if I do not have A? I am a little bit confused...

b = 13, c = 11, A = 31 degrees.

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

You aren't missing anything.

-Dan