Find the hypotenuse of a right triangle on the unit circle with one side equal to $ frac{1}{sqrt{2}} $

In a right triangle on the unit circle, the hypotenuse is always 1. If one side is $ \frac{1}{\sqrt{2}} $, the other side must also be $ \frac{1}{\sqrt{2}} $ to satisfy the Pythagorean theorem:

$ a^2 + b^2 = c^2 $

Here, $ a = \frac{1}{\sqrt{2}} $ and $ b = \frac{1}{\sqrt{2}} $, and the hypotenuse $ c = 1 $:

$ \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 $

$ \frac{1}{2} + \frac{1}{2} = 1 $

Therefore, the hypotenuse is 1.

In a right triangle on the unit circle, the hypotenuse is always 1. Using the Pythagorean theorem:

$ a^2 + b^2 = 1 $

Given $ a = frac{1}{sqrt{2}} $:

$ left(frac{1}{sqrt{2}}
ight)^2 + b^2 = 1 $

$ frac{1}{2} + b^2 = 1 $

Therefore, $ b $ must be $ frac{1}{sqrt{2}} $.

For a unit circle triangle, the hypotenuse is always 1. If one side is $ frac{1}{sqrt{2}} $, the other side is also $ frac{1}{sqrt{2}} $ to satisfy the Pythagorean theorem.

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