CT or Chit-Chat?

[~kitty~]

I put it in CT first, since it's most likely to be moved to the correct place with all these "high-leveled" thinkers, if it doesn't really belong here. I don't think that's abusing, but oh well... here it goes:


What would you say about a math problem that is like this?

---In a baseball diamond, where each side is 90 ft in length, using the laws of Sine and Cosine, how will you determine the distance from the Home base to 3rd base is the pitchers mound is 60.5 ft from Home? Solve and Show all steps neatly.


Tell me, is that a good question? I don't think it is... I think it's a question that assumes you know things that aren't quite related to the question itself.

For me, I got the distance to be over 100... I don't think that's possible. What do you guys get?


Quick Edit: The question is paraphrased, but I gave all the data used in the problem in the sentence.

[gnr61]

nice try haha

[~kitty~]

Originally Posted by Saik0Shinigami

It's abusing because you're trying to make us do you're homework for you ;P


The question is poor because it's missing information. It's not making you assume information, it's plain missing information. There's no information on where the mound is actually located.

And the greatest distance possible is actually something like 89.99(continuing) but that's not going to be your answer.

oh and just to show what you did wrong(because I know what you did don't lie), it said home base to 3rd... Not home base to first, to second then to third.

Okay, first thing... that came from a quiz, not homework.


Second thing, I gave you everything given to me that was on that question.


Here's how it goes. The pitchers mound is in the middle of the baseball diamond, since if you know about baseball that would be most likely an obvious one to you (which makes it a dumb question for a quiz), and 3rd base is just to the left side (which you do not skip to if you play baseball) and the question never stated it was a person doing it. It only asked how much distance was in between them.

That makes me realize something, and my bad... I meant distance between the pitchers mound and 3rd base. That still doesn't change the fact that in this problem, you do need to find the distance between 3rd and Home.

It's 90 ft, actually. Home to 3rd is 90 ft. Home to Pitchers Mound is 60.5 ft. Angle Pitcher/Home/Third is 45 degrees. Now you are given two sides and one angle. This is now where you use the law of Sines/Cosines.

Now tell me, how is this a good question to put on a quiz? I don't see how it can be.

[Squeek]

First of all, we have a forum for this. It's called Homework help. I've moved your thread there.

Second of all, if you're saying it's a baseball diamond with all sides equal to 90 feet, then why isn't the distance between home and third...90 feet?

Edit: Oh I see you just mucked up the wording. Next time, just quote the problem word for word.

[~kitty~]

Originally Posted by Squeek

First of all, we have a forum for this. It's called Homework help. I've moved your thread there.

Second of all, if you're saying it's a baseball diamond with all sides equal to 90 feet, then why isn't the distance between home and third...90 feet?

Edit: Oh I see you just mucked up the wording. Next time, just quote the problem word for word.

I don't have the quiz with me D:

Originally Posted by Saik0Shinigami

Law of Sines
(sin A)/a = (sin B)/b = (sin C)/c
Law of Cosines
c2 = a2 + b2 – 2ab cos C
a = catcher to mound side
b = 3rd to mound
c = catcher to 3rd

angles are the opposite of the sides

The data to start...
a = 60.5
b = ??
c = 90

A = ??
B = 45
C = ??

Law of cosines to find the third side.
b² = a² + c² - 2ac cosB (changed to fit our needs)
b² = 60.5² + 90² - 2(60.5)(90) cos(45)
b² = 3660.25 + 8100 - 10890 (.5253219888[use scientific calc for actual value])
b² = 6039.493542
Sqrt both sides
b = 77.71417851
Most teachers will accept 77.7 ft

This is enough to actually get your answer of 167.7 ft (90+77.7)
but I think she wanted you to solve the WHOLE triangle, cause that's how all math teachers are...

Here's where sine comes in

Sin A/a = Sin B/b

Sin A/60.5 = Sin45/77.71417851
Sin A/60.5 = .0109491413
multiply both sides
Sin A = .662423051
then you arcsin to find for A
A = 41.48[...]
Just round to 41.5

Then use 180 - 41.5 - 45 = C(all triangles must = 180 degrees)
C = 93.5

That gives you every value back... from there you can hand her whatever information she wants...

a= 60.5 ft
b= 77.7 ft
c= 90 ft

A= 41.5 degrees
B= 45 degrees
C= 93.5 degrees



If you're using a graphing calculator make sure your not in Radian mode... I almost went bezerk on mine a few minutes ago because it was returning back A = .77[...]


Final note... the pitchers mound isn't "in the center" of the diamond... it's actually a little "south" of the center.

If anyone ninja'd me I'm going to f***ing kill you and your children.

My point is more that she even used the question in the first place rather than me looking for the answer.