6.2 – Law of Cosines
Jan 10th, 2011 by corricelli
Hello,
Use this section to post questions/ respond to questions in Chapter 6, Section 2.
Thanks,
Mrs. Corricelli
An online math community
Jan 10th, 2011 by corricelli
Hello,
Use this section to post questions/ respond to questions in Chapter 6, Section 2.
Thanks,
Mrs. Corricelli
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Misty Look by Sadish
Dear fellow bloggers,
On number 51 on the homework, the problem involving a baseball diamond, two side lengths of the triangle are the only values given. I assume this means it is safe to assume that the angle from the pitchers mound is 90 degrees. If this is where my mistake is please let me know. But then after that I just used the ole pythagorean theorum but my value is off by about 3
joejoe,
i think you have to assume that the baseball diamond is a square..which is king of ironic but that’s because it says each side is 90 feet.
then, you know the angle across from the side your trying to find is 45 degrees because the line from the pithcers mound is just straight down
with that you can use the law of cosines, woo
kind*
Hey Teams 2 and 7:
Check out this link regarding the “baseball diamond is a square” discussion: http://www.geom.uiuc.edu/~demo5337/Group3/bball.html.
Note, though, not every equilateral quadrilateral is a square. It could be a rhombus. It is interesting, too, that they use the word “diamond” which sorta implies rhombus. I think it should be called a baseball square… Or is there a baseball diamond that is not a square? I could not find one in the brief search I did? Anyone?
Keep me posted!
Mrs. Corricelli
Mrs. Corricelli,
Are we going to need to know Heron’s area formula for the quiz on Tuesday? Because it was in some of the homework problems, but we never really used it in class.
Yes! It is a good thing to know. You will not need it for the midterm.
Thanks,
Mrs. Corricelli