Question

If A and B are both angles that terminate in quadrant two
Sin (A) = 7/25 and Cos (B) = -3/5, Find Sin(A+B)

Answer 1

hello?

Answer 2

To do this you can use the angle sum formula:\[\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)\]
There are for elements to the right side of this equation, and you already have the values for two of them: sin(A) and cos(B).
Now you just need to find the other two: cos(A) and sin(B).

To find these values, you use your knowledge of what sin and cos represent. Sin represents the ratio of opposite/hypotenuse of a right triangle. Since sin(A) is 7/25, you know the triangle has 7 for the opposite side and 25 for the hypotenuse. Just use the Pythagorean theorem to figure out the 3rd side, and then with all 3 sides known you can figure out what cos(A) is.\[a^2+b^2=c^2 \]\[b=\sqrt{c^2-a^2}\]\[b = \sqrt{25^2-7^2} = 24\]
So now you know that cos(A) is -24/25 (it's negative because you're told that the angle terminates in quadrant 2, which means it has a negative cos).

Do the same process for B to know that sin(B) = 4/5. Now just plug in all the values:\[\sin(A+B) = (7/25)(-3/5) + (-24/25)(4/5) = -117/125\]

Answer 3

wait, why is it -24?