Final answer:
To find sin ACB in triangle ABC with AB = BC = 25 and AC = 40, we can use the Law of Cosines to find the cosine of angle ACB and then use the relationship sin^2(ACB) = 1 - cos^2(ACB) to find sin(ACB). Plugging in the given values, we find that sin(ACB) is 0.6.
Step-by-step explanation:
To find κACB, we can use the Law of Cosines, which states that in a triangle with sides a, b, and c, and included angle C, the following relationship holds: c^2 = a^2 + b^2 - 2ab*cos(C).
In this case, a = AB = BC = 25, b = AC = 40, and c = ACB (the angle we want to find). Plugging these values into the equation, we have:
ACB^2 = 25^2 + 40^2 - 2*25*40*cos(ACB)
625 = 625 + 1600 - 2000*cos(ACB)
0 = 1600 - 2000*cos(ACB)
cos(ACB) = -1600/2000 = -0.8
Therefore, sin(ACB) = sqrt(1 - cos^2(ACB)) = sqrt(1 - (-0.8)^2) = sqrt(1 - 0.64) = sqrt(0.36) = 0.6