Question

In an isosceles $△ABC$ if AC=BC and $A{B}^2=2A{C}^2$ then $\angle C$ is equal to

• A. ${40}^{\circ }$
• B. ${60}^{\circ }$
• C. ${30}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

Answer: (D)

${90}^{\circ }$

In $△ABC$

$(AB{)}^2=2(AC{)}^2$

$(AB{)}^2=(AC{)}^2+(AC{)}^2$

$\because AC=BC$

$(AB{)}^2=(AC{)}^2+(BC{)}^2$

$\therefore \angle C={90}^{\circ }$

And $△ABC$ is a right angle triangle