Question

If $0<A<B<\pi ,sinA+sinB=\sqrt{\frac{3}{2}}$ and $cosA+cosB=\frac{1}{\sqrt{2}},$ then $A=$

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

Answer 1

Answer: (A)

${\left(sinA+sinB\right)}^2+{\left(cosA+cosB\right)}^2=\frac{3}{2}+\frac{1}{2}=\frac{4}{2}$
$2+2\left(sinAsinB+cosAcosB\right)=2$
$cos\left(B-A\right)=0\Rightarrow B-A=\frac{\pi }{2}\Rightarrow B=\frac{\pi }{2}+A$
$cosA+cosB=\frac{1}{\sqrt{2}}\Rightarrow cosA+cos\left(\frac{\pi }{2}+A\right)=\frac{1}{\sqrt{2}}$
$\Rightarrow cosA-sinA=\frac{1}{\sqrt{2}}\Rightarrow cosA\cdot \frac{1}{\sqrt{2}}-sinA\frac{1}{\sqrt{2}}=\frac{1}{2}$
$sin\left(45^o-A\right)=\frac{1}{2}=sin30^o\Rightarrow 45^o-A=30^o$
$\Rightarrow A=15^o$