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Solve for missing side length

Solve for missing side length-example-1

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Explanation:

imagine this triangle to be a trigonometric triangle in a circle.

the angle is at the triangle vertex at the center of the circle.

13×sqrt(2) is then the cosine of 45° multiplied by the Hypotenuse (= radius of the circle) x.

13×sqrt(2) = cos(45)×x

x = 13×sqrt(2)/cos(45) = 26

y = sin(45)×x = 13×sqrt(2)×sin(45)/cos(45) =

= 13×sqrt(2)×tan(45) = 13×sqrt(2)×1 = 18.38477631...

as expected, a right-angled triangle with 45° base angle must be an isoceles triangle (both legs are equally long).

y = 13×sqrt(2)

x = 26

so, if you are not supposed to use trigonometric functions yet, we can wrap it up in the opposite way :

because it is a right-angled triangle with a base angle of 45° (which makes also the other base angle 45°), it is an isoceles triangle, and therefore

y = 13×sqrt(2)

x we get then via Pythagoras :

x² = (13×sqrt(2))² + (13×sqrt(2))² = 169×2 + 169×2 =

= 4×169 = 676

x = sqrt(676) = 26

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