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Welp
 one year ago
http://i.imgur.com/3dAFKoJ.png
since it looks like an equilateral triangle but it's not so I have no idea sure if I should use sin or cos.
Welp
 one year ago
http://i.imgur.com/3dAFKoJ.png since it looks like an equilateral triangle but it's not so I have no idea sure if I should use sin or cos.

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Welp
 one year ago
Best ResponseYou've already chosen the best response.0"Find the area of the triangle"

Welp
 one year ago
Best ResponseYou've already chosen the best response.0Forgot to add that in there

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3It's not equilateral as the sides are not congruent. We know that \[A = \frac{ 1 }{ 2 }bh\] but this is for a right angle triangle dw:1438328568192:dw so using this as a reference to draw a new picture

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438328622099:dw so for a non right angle triangle we can use the area \[A = \frac{ 1 }{ 2 }absinC\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438328714694:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Now you just have to plug and chug.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3\[A = \frac{ 1 }{ 2 }acsinB\] so for you use this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This isn't equilateral triangle. An equilateral triangle has 60 degrees angle on each sides
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