In ∆ A B C , A = 75 ° , B = 45 ° then the value of b + c 2 =

Given in $∆ABC, A=75°,B=45°$

$\Rightarrow C=180°-75°-45°=60°$ (Using Angle Sum property of triangle)

Also, $\sin 75°=\frac{\sqrt{3}+1}{2\sqrt{2}}$, $\sin 45°=\frac{1}{\sqrt{2}}$ and $\sin 60°=\frac{\sqrt{3}}{2}$

Now, using Sine rule, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\Rightarrow \frac{a}{{\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}}}=\frac{b}{{\displaystyle \frac{1}{\sqrt{2}}}}=\frac{c}{{\displaystyle \frac{\sqrt{3}}{2}}}$

$\Rightarrow \frac{\left(2\sqrt{2}\right)a}{\sqrt{3}+1}=\sqrt{2}b=\frac{2c}{\sqrt{3}}$

$\Rightarrow b=\frac{2\sqrt{2}a}{\sqrt{2}\left(\sqrt{3}+1\right)}=\frac{2a}{\sqrt{3}+1}$

and $c=\frac{\left(2\sqrt{2}\right)a}{\sqrt{3}+1}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{6}a}{\sqrt{3}+1}$

Now, $\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{b}+\mathrm{c}\sqrt{2}$

$=\frac{2a}{\sqrt{3}+1}+\frac{\sqrt{6}a}{\sqrt{3}+1}\times \sqrt{2}$

$=\frac{2a+2\sqrt{3}a}{\sqrt{3}+1}$

$=\frac{2a\left(1+\sqrt{3}\right)}{\sqrt{3+1}}=2a=\mathrm{R}.\mathrm{H}.\mathrm{S}.$