The sides $a, b, c$ of a triangle satisfy the relations $c^2=2ab$ and $a^2+c^2=3b^2.$ Then the measure of $\angle BAC$, in degrees, is

 It is given that the sides of triangle, $a, b$ and $c$ satisfy the following relations

$c^2=2ab$
and $a^2+c^2=3b^2$
From Eqs. (i) and (ii), we get

$a^2+2ab=3b^2$

$\Rightarrow a^2+2ab+b^2=4b^2$

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

$\Rightarrow a=b,80c=\sqrt{2}a$