MHB Triangle ABC has area 25*sqrt(3). if Angle BAC=30 degrees, find |AC|=|BC|=?

[Elissa89]

Triangle ABC has area 25*sqrt(3). if Angle BAC=30 degrees, find |AC|=|BC|=?

the answer I got was 10*4th root(3)

Is this correct?

I am asking because someone other than my professor wrote the study guid for us for the final and I am not 100% sure what |AC|=|BC| means as my professor never used it. I'm assuming it means side lengths of the triangle.

 

[skeeter]

Triangle ABC has area 25*sqrt(3). if Angle BAC=30 degrees, find |AC|=|BC|=?

the answer I got was 10*4th root(3)

Is this correct?

I am asking because someone other than my professor wrote the study guid for us for the final and I am not 100% sure what |AC|=|BC| means as my professor never used it. I'm assuming it means side lengths of the triangle.

the triangle is isosceles with $m\angle C = 120^\circ$ and $a = BC = b = AC$

$Area = \dfrac{1}{2}ab\sin(C)$

$25\sqrt{3} = \dfrac{1}{2}a^2 \sin(120^\circ)$

try again to solve for $a$