Question

If \( \cos(A - B) = \frac{\sqrt{3}}{2} \) and \( \sin(A + B) = \frac{\sqrt{3}}{2} \), where \( 0^\circ<(A + B) \leq 90^\circ \) and \( A>B \), then find the values of \( A \) and \( B \).

Hint:

When solving for angles using trigonometric functions, use known values of trigonometric ratios for common angles (like \( 30^\circ, 45^\circ, 60^\circ \)) to simplify the solution.

Answer 1

We are given the equations: \[ \cos(A - B) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(A + B) = \frac{\sqrt{3}}{2}. \] 1. Solve for \( A - B \):
From \( \cos(A - B) = \frac{\sqrt{3}}{2} \), we know that: \[ A - B = 30^\circ \quad \text{(since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \))}. \] 2. Solve for \( A + B \):
From \( \sin(A + B) = \frac{\sqrt{3}}{2} \), we know that: \[ A + B = 60^\circ \quad \text{(since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \))}. \] 3. Solve the system of equations:
Now we have the system of equations: \[ A - B = 30^\circ \quad \text{and} \quad A + B = 60^\circ. \] Adding these two equations: \[ 2A = 90^\circ \quad \Rightarrow \quad A = 45^\circ. \] Substitute \( A = 45^\circ \) into \( A + B = 60^\circ \): \[ 45^\circ + B = 60^\circ \quad \Rightarrow \quad B = 15^\circ. \]
Conclusion: The values of \( A \) and \( B \) are \( A = 45^\circ \) and \( B = 15^\circ \).