Asked by Alexis
Find the area of a triangle if the sides of the triangle have the lengths 849mi, 930mi, and 1301mi.
Here's my work:
a=849
b=930
c=1301
A=.5(849)(930)sinA
sin A = 394,785 mi^2
(this number doesn't seem right to me, can you explain to me what I did wrong)?
Here's my work:
a=849
b=930
c=1301
A=.5(849)(930)sinA
sin A = 394,785 mi^2
(this number doesn't seem right to me, can you explain to me what I did wrong)?
Answers
Answered by
Damon
http://www.mathopenref.com/heronsformula.html
Answered by
Alexis
@Damon, thank you! :)
Answered by
Damon
You are welcome.
Answered by
Steve
how did you get sinA?
Also, the area = (bc sinA)/2, not (ab sinA)/2
849^2 = 930^2 + 1301^2 - 2*930*1301 cosA
A = 40.6°
area = b*c*sinA/2 = 930*1301*0.651/2 = 393,879
Also, the area = (bc sinA)/2, not (ab sinA)/2
849^2 = 930^2 + 1301^2 - 2*930*1301 cosA
A = 40.6°
area = b*c*sinA/2 = 930*1301*0.651/2 = 393,879
Answered by
Alexis
@Steve, I plugged in the numbers in the equation. I see where I went wrong. Thank you!
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