Please Assist, been stuck quite some time on this

Question

Sorry to bother but I can't seem to wrap my head around how I should be doing this problem.

Two Angles A and B are in the same quadrant. SinA= -4/5 and TanB= 1/5.

Find

A) sin(a+b)

b) cos(a+b)

c) The quadrant containing a+b

Answer 1

Sine is negative in Q3 and Q4

Tangent is positive in Q1 and Q3

Therefore, angles A and B are in Q3 and a sketch would look like this:

Using the Pythagorean Theorem on the left picture (angle A) we get x = -3 meaning cos(A) = -3/5

Using the Pythagorean Theorem on the right picture (angle B) we get h = √26 meaning sin(B) = -1/√26 and cos(B) = -5/√26

Using the angle addition identity for sine:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B) = (-4/5)(-5/√26) + (-3/5)(-1/√26) = 4/√26 + 3/(5√26) = 20/(5√26) + 3/(5√26) = 23/(5√26) = 23√26/(5•26) = 23√26/130

cos(A + B) = cos(A)cos(B) – sin(A)sin(B) = (-3/5)(-5/√26) - (-4/5)(-1/√26) = 15/(5√26) - 4/(5√26) = 11/5√26) = 11√26/(5•26) = 11√26/130

Since sine and cosine are both positive in Q1 (and these answers are both positive), then A+B is in Q1.