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Mathematics 18 Online
OpenStudy (sleepyjess):

Law of cosines help?

OpenStudy (sleepyjess):

Find the area of the triangle with the given measurements. B = 105, a = 12 cm, c = 19 cm

OpenStudy (studygurl14):

Okay, so do you know the law of cosines formulas?

OpenStudy (sleepyjess):

Yes

OpenStudy (studygurl14):

You're going to need to use Heron's formula for this.

OpenStudy (sleepyjess):

Yep, I need to find the side length b right?

OpenStudy (studygurl14):

Correct.

OpenStudy (studygurl14):

Use law of cosines first to find b. Then use Heron's formula.

OpenStudy (sleepyjess):

Ok, I get Heron's area formula, but I'm not sure how to get the 3rd side length.

OpenStudy (danjs):

\[b^2 = 12^2 + 19^2-2*12*19*\cos(105)\]

OpenStudy (studygurl14):

Law of cosines: \(\large b^2=a^2+c^2-2ac\cos B\) Plug in and solve for b.

OpenStudy (sleepyjess):

\(\sf b^2~=~12^2~+~19^2~-~2ac~cosB\)

OpenStudy (studygurl14):

You know B.

OpenStudy (studygurl14):

you need to do a little more plugging in

OpenStudy (sleepyjess):

\(\sf b^2~=~12^2~+~19^2~-~2ac~cos105\)

OpenStudy (studygurl14):

Not quite \(\large b^2=(12)^2+(19)^2-2(12)(19)\cos (105)\)

OpenStudy (sleepyjess):

Oh whoops :/

OpenStudy (studygurl14):

:) Can you do the rest, or no?

OpenStudy (sleepyjess):

I think so, I will let you know what I come up with if that is ok?

OpenStudy (studygurl14):

Okay,

OpenStudy (sleepyjess):

I got approximately 25 for b

OpenStudy (studygurl14):

yes, although I prefer not to round too much until the very end, so I used 24.96 to plug in for the Heron's formula

OpenStudy (sleepyjess):

Ok :)

OpenStudy (studygurl14):

Do you need help for the heron's part, or no...?

OpenStudy (sleepyjess):

No, I had to find a notebook haha xD

OpenStudy (studygurl14):

okay, let me know what you get.

OpenStudy (sleepyjess):

110.11

OpenStudy (studygurl14):

Perfect. Nice job. :)

OpenStudy (sleepyjess):

yay! thank you :)

OpenStudy (studygurl14):

You're welcome. :)

jimthompson5910 (jim_thompson5910):

There's no need to use the law of cosines since there is a shorter way Look on this link where it says "Knowing Two Sides and the Included Angle" http://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html however, your method works just fine

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