Question

If $0 < A < B < \pi , \, sin A+sin⁡B=\sqrt{\frac{3}{2}}$ and $cos A+cos ⁡ B=\frac{1}{\sqrt{2}},$ then $A=$

• A. $15^{o}$
• B. $30^{o}$
• C. $45^{o}$
• D. $22\frac{1}{2}^{o}$

Answer 1

Answer: (A)

$\left(sin A + sin ⁡ B\right)^{2}+\left(cos ⁡ A + cos ⁡ B\right)^{2}=\frac{3}{2}+\frac{1}{2}=\frac{4}{2}$
$2+2\left(sin A sin ⁡ B + cos ⁡ A c o s B\right)=2$
$cos\left(B - A\right)=0\Rightarrow B-A=\frac{\pi }{2}\Rightarrow B=\frac{\pi }{2}+A$
$cos A+cos ⁡ B=\frac{1}{\sqrt{2}}\Rightarrow cos ⁡ A+cos\left(\frac{\pi }{2} + A\right)=\frac{1}{\sqrt{2}}$
$\Rightarrow cos A-sin ⁡ A=\frac{1}{\sqrt{2}}\Rightarrow cos ⁡ A\cdot \frac{1}{\sqrt{2}}-sin ⁡ A\frac{1}{\sqrt{2}}=\frac{1}{2}$
$sin\left(4 5^{o} - A\right)=\frac{1}{2}=sin 3 0^{o}\Rightarrow 45^{o}-A=30^{o}$
$\Rightarrow A=15^{o}$