Question

In an isosceles $\Delta ABC$, if $AC=BC$ and ${AB}^2=2{AC}^2$, then $\angle C$ is equal to :

• A. ${45}^o$
• B. ${60}^o$
• C. ${30}^o$
• D. ${90}^o$

Answer 1

Answer: (D)

${90}^o$

Given AC = BC and ${\left(\frac{AB}{AC}\right)}^2=2⟹{AB}^2=2{AC}^2$
If AC = BC, then angles opposite these two sides must also be equal. If these angles are equal to 60 degrees, then the triangle would be an acute angled and equilateral triangle, which clearly is not the case.
Since ${AB}^2=2{AC}^2$, it must be a right triangle. Hence angle C = 90 degrees.
${AB}^2={BC}^2+{AC}^2⟹{AB}^2={BC}^2+{AC}^2⟹{AB}^2={BC}^2+B{C}^2{⟹AB}^2=2{AC}^2[BC=AC]$