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x^2 + (x+2)^2 = (2√5)^2 z Tw. Pitagorasa
x^2 + x^2 + 4x + 4 = 4*5
2x^2 + 4x + 4 = 20
2x^2 + 4x - 16 = 0
x^2 + 2x - 8 = 0
∆=4 + 4*8 = 4 + 32 = 36
√∆ = 6
x1 = (-2 - 6)/2 = -4 - nie spelnia warunkow zadania.
x2=(-2 + 6)/2 = 2
Zatem przyprostokatne maja dlugosc: 2, 4
tga=4/2=2 (gdzie a jest katem naprzeciw boku o dl. 4)