Suppose sinA = 12/13 with 90º≤A≤180º. Suppose also that sinB = -7/25 with -90º≤B≤0º. Find cos(A - B).

To find cos(A - B), we can use the trigonometric identity:

cos(A - B) = cosA * cosB + sinA * sinB

First, let's find the values of cosA and cosB.

Given that sinA = 12/13, we can use the Pythagorean identity to find cosA:

cos^2(A) + sin^2(A) = 1
cos^2(A) + (12/13)^2 = 1
cos^2(A) = 1 - (12/13)^2
cos^2(A) = 1 - 144/169
cos^2(A) = (169 - 144) / 169
cos^2(A) = 25/169
cos(A) = ±√(25/169) = ±(5/13)

Since 90º ≤ A ≤ 180º, we take the negative square root:

cos(A) = -(5/13)

Given that sinB = -7/25, we can use the Pythagorean identity to find cosB:

cos^2(B) + sin^2(B) = 1
cos^2(B) + (-7/25)^2 = 1
cos^2(B) + 49/625 = 1
cos^2(B) = 1 - 49/625
cos^2(B) = (625 - 49) / 625
cos^2(B) = 576/625
cos(B) = ±√(576/625) = ±(24/25)

Since -90º ≤ B ≤ 0º, we take the positive square root:

cos(B) = 24/25

Now that we have the values of cosA and cosB, we can substitute them into the formula for cos(A - B):

cos(A - B) = cosA * cosB + sinA * sinB
= -(5/13) * (24/25) + (12/13) * (-7/25)
= -(120/325) - (84/325)
= -204/325

Therefore, cos(A - B) = -204/325.