Doubt regarding the quadrant of angles This question is given in a text book.

and the answer given is

My doubt is that when $A > 0$ and  $sin$ $A$  is positive, the angle $A$ may be in the first or second quadrant. It results in two values of $A$.
Then, how can we determine the sign of $cos$ $A$?
Similarly, angle $B$ is less than $\pi/2$ and $cos A$ is positive, it may be in first or 4th quadrant.
 How can we determine the sign of $sin$ $B$?
We have to use the formula of $sin$$(A+B)$.
 A: Since it's given that $0 \lt A, B \lt \frac{\pi}{2}$, you have as Andrew Chin's question comment states, i.e., $0 \lt A \lt \frac{\pi}{2}$ and $0 \lt B \lt \frac{\pi}{2}$. This means $0 \lt \sin(A), \cos(A), \sin(B), \cos(B) \lt 1$, i.e., all of these values are always positive. Since the $4$ formulas used in the question can be expressed using just those $4$ trig. values (i.e., $\sin(A), \cos(A), \sin(B), \cos(B)$), you don't have to worry about the signs of each component as it's always positive. Thus, from $\sin^2(x) + \cos^2(x) = 1$ for all $x$, you have $\sin(A) = \frac{4}{5} \implies \cos(A) = \frac{3}{5}$ and $\cos(B) = \frac{5}{13} \implies \sin(B) = \frac{12}{13}$, so for example with part (i) of your question you get 
$$\begin{equation}\begin{aligned}
\sin(A + B) & = \sin(A)\cos(B) + \cos(A)\sin(B) \\
& = \left(\frac{4}{5}\right)\left(\frac{5}{13}\right) + \left(\frac{3}{5}\right)\left(\frac{12}{13}\right) \\
& = \frac{20}{65} + \frac{36}{65} \\
& = \frac{56}{65}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
