Question

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

Answer 1

if you mean a is the side opposite point A and b the side opposite point B:

draw the triangle then extend side b past point C. Now draw a line perpendicular to the extended side b starting at point B. Now you have a 45-45-90 right triangle with hypotenuse BC=8. So the base and height of that triangle is 8/sqrt(2). Now find the Area of the big right triangle with hypotenuse BA. This is:
(1/2)*(9+8/sqrt(2))*(8/sqrt(2))=36/sqrt(2)+32/sqrt(2)=68/sqrt(2)

Area of smaller right triangle hypotenuse BC.

(8/sqrt(2))^2*1/2=16

subtract the smaller area from the total to find the area you want.

68/sqrt(2)-16=34sqrt(2)-16

A=34sqrt(2)-16