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Finding the Degree of of triangle.

  • Thread starter anto3232
  • Start date
1. Homework Statement
I have to find the degree at which im launching a water balloon and i have 2 quadratic formulas for the X and Y values versus time. The X and Y quadratic formulas for it was, X: 55.81t^2+(-2.197t)+0.5222
Y:52.62t^2+(-2.358t)+.4135


2. Homework Equations
Is A the length of x or y, and for 45 degrees shouldnt it be equal? for it to equal 45 degrees?


3. The Attempt at a Solution
Well i used a^2+b^2=C^2 and got 76.70, then i figured the angle of that side was 90 degrees so i set Sin (90)/76.70=sin (b)/52.62
and got .61 so i converted to % which is 61% but looking at the graph it didnt make much sense seeing as the length of A is greater than the Heigh (b) so wouldnt angle B, be smaller ?
 
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1. Homework Statement
I have to find the degree at which im launching a water balloon and i have 2 quadratic formulas for the X and Y values versus time. The X and Y quadratic formulas for it was, X: 55.81t^2+(-2.197t)+0.5222
Y:52.62t^2+(-2.358t)+.4135


2. Homework Equations
Is A the length of x or y, and for 45 degrees shouldnt it be equal? for it to equal 45 degrees?
This space is for relevant equations, not questions. Anyway, how do you know that the balloon was launched at an angle of 45 degrees?
3. The Attempt at a Solution
Well i used a^2+b^2=C^2 and got 76.70,
For what, c? And what did you use for a and b?
then i figured the angle of that side was 90 degrees so i set Sin (90)/76.70=sin (b)/52.62
and got .61
For what? Help us out here. Is this sin(b)?
so i converted to %
Why did you do that? I have no idea why you'd want to convert to a percent.
which is 61% but looking at the graph it didnt make much sense seeing as the length of A is greater than the Heigh (b) so wouldnt angle B, be smaller ?
You need to show us how you got sides a and b of your triangle.
 
This space is for relevant equations, not questions. Anyway, how do you know that the balloon was launched at an angle of 45 degrees?
Thats what im trying to figure out, im pretty sure it didnt because for my y graph i got 52.62 and for the x graph i got 55.81 which isnt equal meaning it aint 45 Degrees, so what im trying to say is since i know the a and b how do i figure out the angle?
For what, c? And what did you use for a and b?
a=the x which is 55.81 and b= the y which is 52.62.
For what? Help us out here. Is this sin(b)?

Why did you do that? I have no idea why you'd want to convert to a percent.


You need to show us how you got sides a and b of your triangle.
Well i thought A for some reason was the 55.81 and B was the 52.62 from the two quadratic formulas, but i think that isnt right...

Hmm hold on ill lemme take a look at what im doing
 
okay so i have a water balloon launcher, it starts off at 0.51410733857,0.404340993843
and at the end of the launch its at 1.2249085093,1.03940105622, this is in meters btw..
What i need to do is find the angle or degree it launch's at. Do i X2-X1/Y2-Y1 it? I dont know that would just give me the slope but could that help me get the angle?
 
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4,301
What's the exact wording of the problem? It seems like the problem as you first described it is different from what you have now. Where did the quadratic equations come from?
 

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