Need help with precalcuclus will give owl bucks!!

- matlee

Need help with precalcuclus will give owl bucks!!

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- katieb

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- anonymous

whats the question

- anonymous

ok what do u need help with

- matlee

I will post 1 by 1. Ill show .

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## More answers

- anonymous

ok sounds good

- matlee

|dw:1450395303398:dw|

- matlee

both of u can help

- anonymous

But, actually if you can can you please post it in another question not the same tab

- matlee

Y

- anonymous

so that way when you give you medals we get the full credit

- anonymous

Its only fair

- matlee

Ok lets do this question first

- anonymous

ok

- matlee

Im giving owl bucks not really medals

- matlee

Thanks tho

- anonymous

ok

- anonymous

so first a triangle equals 180 all together when added up

- matlee

Yes

- anonymous

And im pretty sure c is a right angle

- anonymous

so that equals 90

- 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

everything is not drawn to scale we can propose

- matlee

|dw:1450395646594:dw|

- rishavraj

u remember
\[\cos A = \frac{ b^2 + c^2 - a^2 }{ 2bc }\]
???????the cosine rule

- anonymous

Ohh ok so its not drawn accurately?

- 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

@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

but yes i guess since we are looking for side A it would be

- matlee

wait would we just use pythroean therom?

- rishavraj

nah u cant use pythagorean theorem ..its not a right triangle

- whpalmer4

No, Pythagorean Theorem only works on right triangles, and this is not a right triangle.

- matlee

Oh i see, thanks

- matlee

So how should i go about this

- rishavraj

@matlee
http://www.askiitians.com/iit-jee-properties-and-solutions-of-triangles/solution-of-triangles/

- anonymous

But, don''t we have to find the other angles

- rishavraj

@sydneymmelin we simply need to apply the cosine rule

- matlee

I think we do. But we can say angle C is 90 cus it can be any number within 1-158

- matlee

Ok cosine rules

- matlee

What is the cosine rule and how should i know when to use it

- anonymous

well really you cant just assume the angles cause it would be inaccurate

- matlee

sorry i meant cant*

- anonymous

oh lol right

- 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

@matlee

##### 1 Attachment

- matlee

Oh thats great lol

- 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

We need to rearrange that to get \(a\) alone on one side. Can you do that algebra?

- matlee

Kind of

- matlee

Ok i see why is cos cus the two side adjacnet and hypotenus

- rishavraj

@matlee why are u considering it to be hypotenuse.....Its NOT a right triangle

- matlee

oh i thought the largest one was the hypotenus

- whpalmer4

In a right triangle, the long side is the hypotenuse, but a non-right triangle does not have a hypotenuse.

- matlee

cosA(2bc)-b2 - c2 = -a2 i dont know how to get a alone after that

- whpalmer4

how about you multiply both sides by -1, and then take the square root?

- rishavraj

u got the value of b,c and A
use calculator to get the value of cosA

- matlee

(square root(-1(cosA(2bc)-b2 - c2) = (square root ) a)

- matlee

cos of 22 i do?

- 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

Thank you how do you d it so fast lol that takes me for ever

- matlee

Ok so now we have that

- whpalmer4

\[\cos(22^\circ) \approx 0.927184\]

- matlee

Ok

- whpalmer4

\[a\approx \sqrt{(17)^2 + (36)^2 -2(17)(36)(0.927184)} = 21.2\]

- 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

Oh tahnk you so is that what a is?

- 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

I see yes lol! I have sent you my feed back well appreciated

- 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

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

Lol, thank you very much.

- matlee

It would be cool if you can convert them

- whpalmer4

The thought is appreciated, and I'll let someone else get a crack at the next one!

- matlee

Ok awesome

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