Question

in an isosceles ∆ABC,If AC=BC and AB2=2AC2 Then angal C=

Answer 1

By this statement of

AB^2 = 2(AC^2)

And we get, AC = AB /[sqrt(2)]

From this we get, AC is always smaller than AB, so there is no case where this triangle can be right angled isosceles triangle.

So now user can draw triangle and can get assume on AB = BC = 1 unit

Therefore, AC = 0.707 units

Now draw an line from point B, that will intersect Side AC at D, such that it is perpendicular.

And also due to property of isosceles triangle it will bisect AC into 2 equal parts

Say DC = 0.353 units

So now Triangle BDC will form right angle with Angle C has an

adjacent side to be DC = 0.353 units

Hypotenuse BC = 1 unit

Cos of Angle C = DC/BC

Cos Inverse (DC/BC) = Angle C

Angle C = 69.32 Degree

Answer 2

Answer:

70 degrees approx

Step-by-step explanation:

By this statement of

AB^2 = 2(AC^2)

And we get, AC = AB /[sqrt(2)]

From this we get, AC is always smaller than AB, so there is no case where this triangle can be right angled isosceles triangle.

So now user can draw triangle and can get assume on AB = BC = 1 unit

Therefore, AC = 0.707 units

Now draw an line from point B, that will intersect Side AC at D, such that it is perpendicular.

And also due to property of isosceles triangle it will bisect AC into 2 equal parts

Say DC = 0.353 units

So now Triangle BDC will form right angle with Angle C has an

adjacent side to be DC = 0.353 units

Hypotenuse BC = 1 unit

Cos of Angle C = DC/BC

Cos Inverse (DC/BC) = Angle C

Angle C = 69.32 Degree