# prove trignometric value of sin45degree geometrically

In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠ C = 45°.

Now, let BC = AB = a.

Applying the Pythagoras Theorem, we get,

AC^2 = AB^2 + BC^2 = a^2 + a^2 = 2a^2,

So, AC = a √2

As per the trigonometric ratio definitions, we have,

sin 45° = (side opposite to angle 45°)/(hypotenuse) =BC/AC = a/ a √2 = 1/ √2