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If $\sin (A - B) = \frac{1}{2}$ and $\tan (A + B) = \sqrt{3}$, $0^{\circ} \le A + B < 90^{\circ}$, $A > B$ then find the values of A and B.
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Solution: $\sin (A - B) = \frac{1}{2} \Rightarrow A-B = 30^{\circ}$ ..........(i) (1/2 Mark)
$\tan (A + B) = \sqrt{3} \Rightarrow A + B = 60^{\circ}$ ..........(ii) (1/2 Mark)
Solving (i) and (ii) to get $A = 45^{\circ}$, $B = 15^{\circ}$ (1 Mark)
$\tan (A + B) = \sqrt{3} \Rightarrow A + B = 60^{\circ}$ ..........(ii) (1/2 Mark)
Solving (i) and (ii) to get $A = 45^{\circ}$, $B = 15^{\circ}$ (1 Mark)