Question

In an isosceles triangle ABC if AC = BC and $AB^{2} = 2AC^{2}$, then $\angle C$ =
(a) $30 \textdegree$
(b) $45 \textdegree$
(c) $90 \textdegree$
(d) $60 \textdegree$

Answer 1

Answer:

∆ABC is a right triangle right angled at C.

Among the given options option (c) is 90° is the correct answer.

Step-by-step explanation:

Given:  

Δ ABC is an isosceles triangle  with AC = BC , and AB² = 2AC²

In Δ ABC ,

AB² = 2AC²

AB² = AC² + AC²

AB² = AC² + BC²                                 [ ∵ AC = BC ]

AB² = AC² + BC²  …………... (1)

AB is the hypotenuse and  Δ ABC is a right angled  

From eqn (1) , ∠C = 90°  

[ By converse of Pythagoras theorem ]

Hence,∆ABC is a right triangle right angled at C.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer =90 degree

Explanation:

here,

AB² = 2 AC²

AB² = AC² + AC²

AB² = AB² + AC²

Therefore, AB = AC

This Pythagoras theorem to reverse pythagoras theorem, this is here,

ΔABC is a{ right angle triangle} .

Angle C = 90°

Hence, angle proved