area = (1/2)ac sinB

25 = (1/2) 8 * 5 * sinB

sinB = 1.25 > 1

there is a mistake

I think something is wrong .

AB = 5cm and BC = 8cm , so the area of △ABC becomes

(1/2) * 5 * 8 * sin(∠ABC) [cm^2] .

So the possible maximum value of it is 20cm^2 , it can not be 25cm^2 .

Area= 1/2 base*height. height = 5 sin theta. Base is 8

Area= 25 = 5*8 sin theta. sin theta= 25/40=5/8=0.625

theta=38.68 degrees. This is angle ABC. AB cos theta=3.9. The remainig of the base = 8-3.9=4.1. Now find the angle ACB. Tan ACB= 5 sin theta/4.1=0.76. Theta=37.23. The angles are ABC=38.68, ACB=37.23 and

BAC=104.09