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).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.