Solving for Angles A and B

We are given two trigonometric equations and constraints for angles A and B:

• Equation 1: $\sin(A - B) = \frac{1}{2}$
• Equation 2: $\cos(A + B) = \frac{1}{2}$
• Constraint 1: $0^\circ < (A + B) \leq 90^\circ$
• Constraint 2: $A > B$

Determining Angle Values

From Equation 1, $\sin(A - B) = \frac{1}{2}$. The principal value for the angle whose sine is $\frac{1}{2}$ is $30^\circ$. Therefore, we can write:

$A - B = 30^\circ \quad (\text{Equation 3})$

From Equation 2, $\cos(A + B) = \frac{1}{2}$. Given the constraint $0^\circ < (A + B) \leq 90^\circ$, the angle whose cosine is $\frac{1}{2}$ in this range is $60^\circ$. Therefore:

$A + B = 60^\circ \quad (\text{Equation 4})$

Solving the System of Equations

Now we have a system of two linear equations:

1. $A - B = 30^\circ$
2. $A + B = 60^\circ$

Add Equation 3 and Equation 4:

$ (A - B) + (A + B) = 30^\circ + 60^\circ $

$ 2A = 90^\circ $

$ A = \frac{90^\circ}{2} $

$ A = 45^\circ $

Substitute the value of A into Equation 4:

$ 45^\circ + B = 60^\circ $

$ B = 60^\circ - 45^\circ $

$ B = 15^\circ $

Verifying Constraints

The calculated values are $A = 45^\circ$ and $B = 15^\circ$. Let's check the constraints:

• $A > B$: $45^\circ > 15^\circ$, which is true.
• $0^\circ < (A + B) \leq 90^\circ$: $A + B = 45^\circ + 15^\circ = 60^\circ$. Since $0^\circ < 60^\circ \leq 90^\circ$, this is also true.

Both values satisfy the given conditions.