Question

in tirangle ABC, a=8 b=9 and m of angle C = 135 what is the area of ABC

Answer 1

\[A=\frac{1}{2}bcsin(A)\]

Answer 2

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

Answer 3

none of the answers seem to fit what i got.

Answer 4

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)
\[= 36*\sqrt{2}/2 = 18\sqrt{2}\]