matlee
  • matlee
Need help with precalcuclus will give owl bucks!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
whats the question
anonymous
  • anonymous
ok what do u need help with
matlee
  • matlee
I will post 1 by 1. Ill show .

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anonymous
  • anonymous
ok sounds good
matlee
  • matlee
|dw:1450395303398:dw|
matlee
  • matlee
both of u can help
anonymous
  • anonymous
But, actually if you can can you please post it in another question not the same tab
matlee
  • matlee
Y
anonymous
  • anonymous
so that way when you give you medals we get the full credit
anonymous
  • anonymous
Its only fair
matlee
  • matlee
Ok lets do this question first
anonymous
  • anonymous
ok
matlee
  • matlee
Im giving owl bucks not really medals
matlee
  • matlee
Thanks tho
anonymous
  • anonymous
ok
anonymous
  • anonymous
so first a triangle equals 180 all together when added up
matlee
  • matlee
Yes
anonymous
  • anonymous
And im pretty sure c is a right angle
anonymous
  • anonymous
so that equals 90
whpalmer4
  • whpalmer4
You can use the law of cosines here. You can't make any assumption about the angles not labeled as having a specific value.
matlee
  • matlee
everything is not drawn to scale we can propose
matlee
  • matlee
|dw:1450395646594:dw|
rishavraj
  • rishavraj
u remember \[\cos A = \frac{ b^2 + c^2 - a^2 }{ 2bc }\] ???????the cosine rule
anonymous
  • anonymous
Ohh ok so its not drawn accurately?
matlee
  • matlee
I honestly dont know anything but the wordp re caluclus in pre caluclus lol. I would love to have teacher explanation or any explanation is fine
whpalmer4
  • whpalmer4
@sydneymmelin yes, you can't conclude that it is a right angle just because it looks sort of like it might a right angle. If it isn't labeled as such, either 90 degrees or with little square, you have to assume that it is not.
matlee
  • matlee
but yes i guess since we are looking for side A it would be
matlee
  • matlee
wait would we just use pythroean therom?
rishavraj
  • rishavraj
nah u cant use pythagorean theorem ..its not a right triangle
whpalmer4
  • whpalmer4
No, Pythagorean Theorem only works on right triangles, and this is not a right triangle.
matlee
  • matlee
Oh i see, thanks
matlee
  • matlee
So how should i go about this
rishavraj
  • rishavraj
@matlee http://www.askiitians.com/iit-jee-properties-and-solutions-of-triangles/solution-of-triangles/
anonymous
  • anonymous
But, don''t we have to find the other angles
rishavraj
  • rishavraj
@sydneymmelin we simply need to apply the cosine rule
matlee
  • matlee
I think we do. But we can say angle C is 90 cus it can be any number within 1-158
matlee
  • matlee
Ok cosine rules
matlee
  • matlee
What is the cosine rule and how should i know when to use it
anonymous
  • anonymous
well really you cant just assume the angles cause it would be inaccurate
matlee
  • matlee
sorry i meant cant*
anonymous
  • anonymous
oh lol right
rishavraj
  • rishavraj
To solve a triangle is to find the lengths of each of its sides and all its angles. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle. The cosine rule is used when we are given either a) three sides or b) two sides and the included angle.
rishavraj
  • rishavraj
@matlee
1 Attachment
matlee
  • matlee
Oh thats great lol
whpalmer4
  • whpalmer4
We need to find the length of side \(a\). We have angle \(A\), and sides \(b\) and \(c\). \[\cos A = \frac{-a^2+b^2+c^2}{2bc}\]
whpalmer4
  • whpalmer4
We need to rearrange that to get \(a\) alone on one side. Can you do that algebra?
matlee
  • matlee
Kind of
matlee
  • matlee
Ok i see why is cos cus the two side adjacnet and hypotenus
rishavraj
  • rishavraj
@matlee why are u considering it to be hypotenuse.....Its NOT a right triangle
matlee
  • matlee
oh i thought the largest one was the hypotenus
whpalmer4
  • whpalmer4
In a right triangle, the long side is the hypotenuse, but a non-right triangle does not have a hypotenuse.
matlee
  • matlee
cosA(2bc)-b2 - c2 = -a2 i dont know how to get a alone after that
whpalmer4
  • whpalmer4
how about you multiply both sides by -1, and then take the square root?
rishavraj
  • rishavraj
u got the value of b,c and A use calculator to get the value of cosA
matlee
  • matlee
(square root(-1(cosA(2bc)-b2 - c2) = (square root ) a)
matlee
  • matlee
cos of 22 i do?
whpalmer4
  • whpalmer4
\[\cos A = \frac{-a^2+b^2+c^2}{2bc}\]\[2 b c \cos A = -a^2+b^2+c^2\] \[2bc \cos A - b^2-c^2 = -a^2\] \[-2bc\cos A + b^2 + c^2 = a^2\]\[a = \sqrt{b^2+c^2-2bc\cos A}\] Take the positive square root.
matlee
  • matlee
Thank you how do you d it so fast lol that takes me for ever
matlee
  • matlee
Ok so now we have that
whpalmer4
  • whpalmer4
\[\cos(22^\circ) \approx 0.927184\]
matlee
  • matlee
Ok
whpalmer4
  • whpalmer4
\[a\approx \sqrt{(17)^2 + (36)^2 -2(17)(36)(0.927184)} = 21.2\]
whpalmer4
  • whpalmer4
And if you want to check it, here's a handy triangle calculator with the data already plugged in: http://www.calculator.net/triangle-calculator.html?vc=&vx=17&vy=&va=22&vz=36&vb=&angleunits=d&x=72&y=7
matlee
  • matlee
Oh tahnk you so is that what a is?
whpalmer4
  • whpalmer4
As you can see from the diagram it makes of the actual triangle, this wasn't anywhere close to being a right triangle!
matlee
  • matlee
I see yes lol! I have sent you my feed back well appreciated
matlee
  • matlee
I am not able to send so much although i have so many questions but i am well grateful for the explanatory help. I will be making a new question. Also more than 1 person can help me if they follow all the way thorugh thank you
whpalmer4
  • whpalmer4
I'm not in this for the money. As I'm fond of saying, a few thousand medals and $3.50 will get me a cup of coffee at Starbucks :-)
matlee
  • matlee
Lol, thank you very much.
matlee
  • matlee
It would be cool if you can convert them
whpalmer4
  • whpalmer4
The thought is appreciated, and I'll let someone else get a crack at the next one!
matlee
  • matlee
Ok awesome

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