If $\sin \left( A-B \right)=\dfrac{1}{2}$ and $\cos \left( A+B \right)=\dfrac{1}{2},{{0}^{\circ }}<{A+B}\le {{90}^{\circ }},A>B$ find A and B.

Hint:We know that $\sin {{30}^{\circ }}=\dfrac{1}{2}$ and $\cos {{60}^{\circ }}=\dfrac{1}{2}$. So, we will replace $\dfrac{1}{2}$ with sin 30˚ first and then we will replace cos 60˚ by $\dfrac{1}{2}$ and we will try to find the relation between A and B. We will get two relations, (A - B) = 30˚ and (A + B) = 60˚. Using these two relations, we will find the values of A and B individually.

Complete step-by-step answer:
It is given in the question that $\sin \left( A-B \right)=\dfrac{1}{2}$ and $\cos \left( A+B \right)=\dfrac{1}{2}$. It is also given that the value of A + B lies between 0˚ and 90˚, that is, $0<\left( A+B \right)\le 90$. It is also given that $A>B$, then we have to find the values of A and B.
We know that $\sin {{30}^{\circ }}=\dfrac{1}{2}$ and $\cos {{60}^{\circ }}=\dfrac{1}{2}$. We will use these two values to find the values of A and B.
We have been given that $\sin \left( A-B \right)=\dfrac{1}{2}$. We know that $\sin {{30}^{\circ }}=\dfrac{1}{2}$. So on replacing $\dfrac{1}{2}$ with sin 30˚ in the first relation, we will get,
$\sin \left( A-B \right)=\sin {{30}^{\circ }}$
We can equate the angles on both sides of the above equation, so we will get,
\[\left( A-B \right)={{30}^{\circ }}.........\left( i \right)\]
Also, we have been given that, $\cos \left( A+B \right)=\dfrac{1}{2}$. We know that $\cos {{60}^{\circ }}=\dfrac{1}{2}$. So on replacing $\dfrac{1}{2}$ with cos 60˚ in the second relation, we will get,
$\cos \left( A+B \right)=\cos {{60}^{\circ }}$
We can equate the angles on both sides of the above equation, so we will get,
$\left( A+B \right)={{60}^{\circ }}.........\left( ii \right)$
Since in the question, we have a condition that $0<\left( A+B \right)\le 90$ and we have got the value of $\left( A+B \right)={{60}^{\circ }}$, we can say that this condition is satisfied. So, we can proceed further.
On adding equation (i) and equation (ii), we will get,
 \[\begin{align}
  & A+B={{60}^{\circ }} \\
 & \underline{A-B={{30}^{\circ }}} \\
 & 2A+0={{90}^{\circ }} \\
 & \therefore A={{45}^{\circ }} \\
\end{align}\]
So, we get the value of A as 45˚.
Now, on substituting the value of \[A={{45}^{\circ }}\] in equation (ii), we will get,
$\begin{align}
  & {{45}^{\circ }}+B={{60}^{\circ }} \\
 & B={{60}^{\circ }}-{{45}^{\circ }} \\
 & B={{15}^{\circ }} \\
\end{align}$
Therefore, the value of A is ${{45}^{\circ }}$ and the value of B is ${{15}^{\circ }}$ also A > B as ${{45}^{\circ }}>{{15}^{\circ }}$.

Note: While solving this question, the students may get confused and may interchange the value of cos 60˚ with the value of cos 30˚ in a hurry. Hence, it is recommended that the students take all the values carefully, in order to avoid any mistakes.Also students should remember the important standard trigonometric angles and formulas for solving these types of questions.