## anonymous 5 years ago in tirangle ABC, a=8 b=9 and m of angle C = 135 what is the area of ABC

1. anonymous

$A=\frac{1}{2}bcsin(A)$

2. anonymous

$A=\frac{1}{2} 8\times 9 \times sin(135)$

3. anonymous

none of the answers seem to fit what i got.

4. anonymous

satellite73 is right, to find the area Area = 1/2 bc sin(A) if we want to use the angle C, the formula become Area = 1/2 ab sin(C) = 1/2 * 8 * 9 sin(135) = 36 sin (135) $= 36*\sqrt{2}/2 = 18\sqrt{2}$