Prove that
Question:

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

(a) (60°, 30°)
(b) (60°, 15°)
(c) (45°, 15°)
(d) (60°, 25°)

 

Solution:

As we know that,

$\sin 30^{\circ}=\frac{1}{2}$

Thus,

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

$\Rightarrow A-B=30^{\circ} \quad \ldots(1)$

and $\cos 60^{\circ}=\frac{1}{2}$

Thus,

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

$\Rightarrow A+B=60^{\circ} \quad \ldots(2)$

Solving (1) and (2), we get

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

Hence, the correct optiom is (c).

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