Law of cosines help?
Find the area of the triangle with the given measurements. B = 105, a = 12 cm, c = 19 cm
Okay, so do you know the law of cosines formulas?
Yes
You're going to need to use Heron's formula for this.
Yep, I need to find the side length b right?
Correct.
Use law of cosines first to find b. Then use Heron's formula.
Ok, I get Heron's area formula, but I'm not sure how to get the 3rd side length.
\[b^2 = 12^2 + 19^2-2*12*19*\cos(105)\]
Law of cosines: \(\large b^2=a^2+c^2-2ac\cos B\) Plug in and solve for b.
\(\sf b^2~=~12^2~+~19^2~-~2ac~cosB\)
You know B.
you need to do a little more plugging in
\(\sf b^2~=~12^2~+~19^2~-~2ac~cos105\)
Not quite \(\large b^2=(12)^2+(19)^2-2(12)(19)\cos (105)\)
Oh whoops :/
:) Can you do the rest, or no?
I think so, I will let you know what I come up with if that is ok?
Okay,
I got approximately 25 for b
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
Ok :)
Do you need help for the heron's part, or no...?
No, I had to find a notebook haha xD
okay, let me know what you get.
110.11
Perfect. Nice job. :)
yay! thank you :)
You're welcome. :)
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
Join our real-time social learning platform and learn together with your friends!