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Mathematics 18 Online
OpenStudy (lsdxmdmaddicthc):

What is the area of the triangle below?

OpenStudy (lsdxmdmaddicthc):

@sleepyjess When you get back

OpenStudy (sleepyjess):

Is there a given right angle or is that it?

OpenStudy (lsdxmdmaddicthc):

Nope, that's it...

OpenStudy (sleepyjess):

Okay, have you learned about Law of Sines, Law of Cosines and Heron's Area Formula?

OpenStudy (lsdxmdmaddicthc):

No, I haven't

OpenStudy (sleepyjess):

I would assume since you haven't that there is a 90 degree angle involved

OpenStudy (lsdxmdmaddicthc):

I was supposed to learn about it but I have been absent

OpenStudy (sleepyjess):

oooohhh so you are supposed to know Law of Sines, Cosines and Heron's?

OpenStudy (lsdxmdmaddicthc):

Yes, I can google it. I don't expect you to teach me that lol ;P

OpenStudy (sleepyjess):

I just got done learning about it haha

OpenStudy (dtan5457):

They are all just a bunch of formulas to remember, not TOO hard.

OpenStudy (sleepyjess):

^ exactly

OpenStudy (dtan5457):

In this problem you just need one law of cosine. Then the heron's formula.

OpenStudy (sleepyjess):

again, not the best at drawing|dw:1428553727066:dw|

OpenStudy (sleepyjess):

|dw:1428553761689:dw|

OpenStudy (sleepyjess):

So law of cosines for finding side a \(a^2 = b^2 + c^2 - 2(b)(c)(cos A)\) pretty simple actually :)

OpenStudy (dtan5457):

Gonna let @sleepyjess explain all that. But seriously, copy down these formulas. No way your gonna remember them in your head (at least for now)

OpenStudy (lsdxmdmaddicthc):

Those formulas are a bit complex...It's amazing that someone actually discovered all these relations and equations....

OpenStudy (sleepyjess):

That's what I thought haha

OpenStudy (sleepyjess):

My lessons actually gave the formulas already changed up to find a missing angle if you want me to put them here

OpenStudy (lsdxmdmaddicthc):

Okay :)

OpenStudy (sleepyjess):

But for this problem, we just need a missing side :)

OpenStudy (sleepyjess):

\(cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\) \(cos B = \dfrac{a^2 + c^2 - b^2}{2ac}\) \(cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\) I saw these and was like, how am I supposed to use these?????? But once you plug in everything, it actually is quite simple :)

OpenStudy (lsdxmdmaddicthc):

Okay, so I need to find the cosine of every angle?

OpenStudy (sleepyjess):

No haha I was just giving you those for future reference we just need to find side a

OpenStudy (lsdxmdmaddicthc):

So what would the first step be?

OpenStudy (sleepyjess):

Using one of the formulas tan put up, \(a^2 = b^2 + c^2 - 2(b)(c)(cos A)\)

OpenStudy (lsdxmdmaddicthc):

Okay so a(sq) = (3*3) + (7*7) - 2(3)(7)(cosA) a(sq) = 58 - 42(cos(55)) a(sq) = 58 - 24.1 a(sq) = 34 a = 5.8

OpenStudy (lsdxmdmaddicthc):

I'll give dtan a medal, too because you both helped me :)

OpenStudy (dtan5457):

Your work above is correct to!

OpenStudy (sleepyjess):

Yay! :)

OpenStudy (lsdxmdmaddicthc):

Awwww yeauhhhhh ! \m/ xD

OpenStudy (dtan5457):

Now you have one remaining step, using the heron's formula. Go give it a try.

OpenStudy (sleepyjess):

Whoops, I totally forgot that we were finding the area haha I think tan can finish you off from here :)

OpenStudy (dtan5457):

Haha, thanks. -Start by adding up all the sides and dividing by 2, to get your value of S in the formula.

OpenStudy (lsdxmdmaddicthc):

So wouldn't that just be S = 180/2? Because the sum of the angles in a triangle = 180 degrees

OpenStudy (lsdxmdmaddicthc):

If the above is true then 8.7 = 90 ? wut

OpenStudy (dtan5457):

there is a solid difference between angles and sides lol

OpenStudy (lsdxmdmaddicthc):

OH!! A B and C are referring to the sides....... Wow I am so tired lol ;P

OpenStudy (lsdxmdmaddicthc):

wowwwwwww

OpenStudy (dtan5457):

you know, it's funny that you we actually could have used a law of sin to get the area directly and not do all this cosine/heron formula but.....bit late now LOL

OpenStudy (dtan5457):

SORRY LOL

OpenStudy (dtan5457):

but it's good to know both ways anyhow

OpenStudy (dtan5457):

so continue with the heron formula.

OpenStudy (lsdxmdmaddicthc):

Haha, nice. Well I like math so it's alright 8) Anyways here's the work : S = A+B+C/2 8.7 = A/2 + 5 ?

OpenStudy (lsdxmdmaddicthc):

Wow. sorry for the complete lack of intelligence here. I am barely awake

OpenStudy (dtan5457):

i'm not sure what you did.. but you simply add up the sides here 5.8+3+7 divide that sum by 2

OpenStudy (lsdxmdmaddicthc):

Here we go for real : 8.7 = A+B+C/2 8.7 = 5.8+7+3/2 8.7 = 15.8/2 8.7 = 7.9 what is going on here lol -_-

OpenStudy (dtan5457):

where are you getting the 8.7 from?

OpenStudy (dtan5457):

s=7.9, that is all.

OpenStudy (lsdxmdmaddicthc):

-facepalm-

OpenStudy (dtan5457):

now i suggest a calculator and just plug everything in to that heron formula..and your done.

OpenStudy (lsdxmdmaddicthc):

Yessir

OpenStudy (dtan5457):

don't worry my eyes are just as redshot as yours. xD

OpenStudy (lsdxmdmaddicthc):

What do you mean by that? ^ lol

OpenStudy (dtan5457):

redshot=so tired your eyes are straining to keep open, hence the fact that they are red.

OpenStudy (lsdxmdmaddicthc):

Yep.

OpenStudy (jhannybean):

|dw:1428555822172:dw| can we also not just drop a perpendicular in there to find the height? Considering \[A=\frac{1}{2}bh\]

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