If sin ( A - B ) = frac{1}{2},cos ( A + B ) = | vedclass

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(A) Given that $\sin ( A - B ) = \frac{1}{2}$.Since $\sin 30^{\circ} = \frac{1}{2}$,we have $A - B = 30^{\circ}$ ......$(1)$Also,given that $\cos ( A

Vedclass · Accepted Answer

$A = 45^{\circ}, B = 15^{\circ}$

Vedclass · Answer

(A) Given that $\sin ( A - B ) = \frac{1}{2}$.Since $\sin 30^{\circ} = \frac{1}{2}$,we have $A - B = 30^{\circ}$ ......$(1)$Also,given that $\cos ( A + B ) = \frac{1}{2}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $A + B = 60^{\circ}$ ......$(2)$Adding equations $(1)$ and $(2)$:$(A - B) + (A + B) = 30^{\circ} + 60^{\circ}$$2A = 90^{\circ}$$A = 45^{\circ}$Substituting $A = 45^{\circ}$ in equation $(2)$:$45^{\circ} + B = 60^{\circ}$$B = 60^{\circ} - 45^{\circ}$$B = 15^{\circ}$Thus,$A = 45^{\circ}$ and $B = 15^{\circ}$.

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

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