math
posted by Anonymous .
In the triangle ABC ,AB=11 cm ,BC=9cm an CA=10 cm..Find area of the triangle ABC

If you don't have access to heron's formula, then you can use law of sines
sin a/9 = sin b/10 = sin c/11
c = 180(a+b)
work on that a while and you will find sin b = .8578
Now, h/9 = sin b
so h = 9 sin b
area = 1/2 * 11 * 9 * .8578 = 42.46
Heron's formula says that
with s = (a+b+c)/2
area = sqrt(15*4*5*6) = 42.43
looks good to me. 
WROOOOOONG