Question

lf $(a+b{)}^2=c^2+$ ab in $a\Delta ABC$ and if $\sqrt{2}(\sin A+\cos A)=\sqrt{3}$ then ascending order of angles $A,B,C$ is?

• A. $A,B,C$
• B. $A,C,B$
• C. $B,A,C$
• D. $C,B,A$

Answer 1

The correct option is A $A,B,C$

$(a+b{)}^2=c^2+ab\Rightarrow a^2+b^2+2ab=c^2+ab$

$\Rightarrow \frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}$ $\Rightarrow cosC=-\frac{1}{2}$ (from cosine Rule)

$\Rightarrow C={120}^o$

Now, $\sin (A+{45}^o)=\frac{\sin A}{\sqrt{2}}+\frac{\cos A}{\sqrt{2}}\Rightarrow \sqrt{2}\sin (A+{45}^o)=\sin A+\cos A$ ....................(1)

Now, $\sqrt{2}(\sin A+\cos A)=\sqrt{3}$

$\Rightarrow \sqrt{2}\times \sqrt{2}\sin (A+{45}^o)=\sqrt{3}\Rightarrow \sin (A+{45}^o)=\frac{\sqrt{3}}{2}(=\sin {60}^o)\Rightarrow A+{45}^o={60}^o\Rightarrow A={15}^o$

Now, ${15}^o+B+{120}^o={180}^o\Rightarrow B={45}^o$

Therefore, Ascending Order is $A,B,C$

So, Answer is $A$