Question:
In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then ∠C = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(d) 90°
Given:
AC = BC
$A B^{2}=2 A C^{2}=A C^{2}+A C^{2}=A C^{2}+B C^{2}$
Applying Pythagoras theorem, we conclude that ∆ABC is right angled at C.
or, $\angle C=90^{\circ}$
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