Given sina= 3/4 and cosB = (-5)/13 and this is Quad II find cos(a+B)

cos(u+v) = cos(u)cos(v) - sin(u)sin(v)

To find cos(a + B), we can use the cosine addition formula:

cos(a + B) = cos(a) * cos(B) - sin(a) * sin(B)

First, let's find the value of sin(a) using the given information.

Given that sin(a) = 3/4, we can deduce that cos(a) = sqrt(1 - sin^2(a)) using the Pythagorean identity sin^2(a) + cos^2(a) = 1.

cos(a) = sqrt(1 - (3/4)^2)
= sqrt(1 - 9/16)
= sqrt(16/16 - 9/16)
= sqrt(7/16) = sqrt(7)/4 (since the square root cancels the square of the denominator)

Now, let's compute the value of cos(a + B) using the formula:

cos(a + B) = cos(a) * cos(B) - sin(a) * sin(B)

cos(a + B) = (sqrt(7)/4) * (-5/13) - (3/4) * (sqrt(1 - (-5/13)^2))

To simplify further, we need to find the value of sqrt(1 - (-5/13)^2).

sqrt(1 - (-5/13)^2)
= sqrt(1 - 25/169)
= sqrt(169/169 - 25/169)
= sqrt(144/169)
= 12/13

Plugging this value back into the original equation:

cos(a + B) = (sqrt(7)/4) * (-5/13) - (3/4) * (12/13)
= (-5sqrt(7) - 36sqrt(3))/(52)

Therefore, cos(a + B) = (-5sqrt(7) - 36sqrt(3))/(52) in Quadrant II.