Question

In Triangle ABC

If $A={75}^0,B={45}^0,$ then prove that $b+c\sqrt{2}=2a.$

Answer 1

$\angle A={75}^o,\angle B={45}^o$

$\Rightarrow \angle C={180}^o-\angle A-\angle B$

$={180}^o-{75}^o-{45}^o$

$={180}^o-{120}^o={60}^o$

Using sin rule

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{C}{\sin C}$

$\Rightarrow \frac{a}{\sin {75}^o}=\frac{b}{\sin {45}^o}=\frac{c}{\sin {60}^o}$

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

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

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

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

None L.H.S $=b+c\sqrt{2}$

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

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

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