If sin(A - B) = 1/2 and cos(A + B) = 1/2 , 0° < A + B < 90° and A > B, then find the values of A and B.

Question

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

Answer 1

\(\sin (A-B) =\frac{1}{2} \)

\(\sin (A-B) =\sin 30^{\circ} \quad\left[\because \sin 30^{\circ}=\frac{1}{2}\right] \)

\(A-B =30^{\circ} \cdots (i)\)

\(\cos (A+B) =\frac{1}{2} \)

\(\cos (A+B) =\cos 60^{\circ} \quad\left[\because \cos 60^{\circ}=\frac{1}{2}\right] \)

\(A+B =60^{\circ} \cdots \text { (ii) }\)

\(\text {Adding (i) and (ii) we get, }\)

\(A-B\,+\,A+B =30+60^{\circ} \)

\(2 A =90^{\circ}\)

\(A=\frac{90^{\circ}}{2} =45^{\circ}\)

Putting the Value of A in eq (i).

\(A-B =30^{\circ} \)

\(45^{\circ}-B =30^{\circ} \)

\(B =45^{\circ}-30^{\circ}=15^{\circ}\)

Hence, \(A=45^{\circ}\) and \(B=15^{\circ}\).

Answer 2

\(\sin (A-B)=\frac{1}{2} \Rightarrow A-B=30^{\circ}\)

\(\cos (A+B)=\frac{1}{2} \Rightarrow A+B=60^{\circ}\)

Solving (i) & (ii) to get \(A=45^{\circ},\) \(B=15^{\circ}\)