# In an triangle, AC = BC and AC^2 = 2AB^2, then what will be angle of C?

Dec 25, 2016

${41.41}^{\circ}$

#### Explanation:

Given $A C = B C , \implies \Delta A B C$ is isosceles.
Given $A {C}^{2} = 2 A {B}^{2}$,
$A {C}^{2} = A {D}^{2} + D {C}^{2}$
$A D = \frac{1}{2} A B , \implies A {D}^{2} = \frac{1}{4} A {B}^{2}$
$\implies 2 A {B}^{2} = \frac{1}{4} A {B}^{2} + D {C}^{2}$
$\implies \frac{7}{4} A {B}^{2} = D {C}^{2} , \implies D C = \sqrt{\frac{7}{4}} A B = \frac{\sqrt{7}}{2} A B$
$\tan x = \frac{A D}{D C} = \frac{\frac{1}{2} A B}{\frac{\sqrt{7}}{2} A B}$
$\tan x = \frac{1}{\sqrt{7}}$
$\implies x = {\tan}^{-} 1 \left(\frac{1}{\sqrt{7}}\right) = {20.705}^{\circ}$

$\implies \angle A C B = \angle C = 2 x = 2 \cdot 20.705 = {41.41}^{\circ}$