Geometry Practice Test 2

In ΔABC,
${\overrightarrow{\mathit{A}\mathit{B}}}^2+{\overrightarrow{\mathit{A}\mathit{C}}}^2$ = ${\overrightarrow{\mathit{B}\mathit{C}}}^2$
From this it is clear that the triangle is a right angle triangle.
Now ${\overrightarrow{\mathit{A}\mathit{B}}}^2+{\overrightarrow{\mathit{A}\mathit{C}}}^2$ = ${\overrightarrow{\mathit{B}\mathit{C}}}^2$
and $\overrightarrow{\mathit{B}\mathit{C}}$ = $\sqrt{2}\overrightarrow{\mathit{A}\mathit{B}}$

${\overrightarrow{\mathit{A}\mathit{B}}}^2+{\overrightarrow{\mathit{A}\mathit{C}}}^2$ = ${\left(\sqrt{2}\overrightarrow{\mathit{A}\mathit{B}}\right)}^2$
${\overrightarrow{\mathit{A}\mathit{B}}}^2+{\overrightarrow{\mathit{A}\mathit{C}}}^2$ = $2{\left(\overrightarrow{\mathit{A}\mathit{B}}\right)}^2$
${\overrightarrow{\mathit{A}\mathit{C}}}^2$ = ${\overrightarrow{\mathit{A}\mathit{B}}}^2$
∠A +∠B+∠C = 180°
90° + x° + x° = 180°
2x = 90°
x = 45°