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Matematyka wokół nas 1. Zeszyt ćwiczeń cz. 2 - ukr

Rozwiązanie 1

Dygresja: Uogólnione twierdzenie Pitagorasa dla ogólnego trójkąta (twierdzenie cosinusów) wygląda następująco: Jeśli w trójkącie o bokach długości a, b i c oznaczymy przez  α  miarę kąta leżącego naprzeciw boku c, to zachodzi następująca równość: a2+b2−2abcosα=c2 Funkcja cosα

Utwórzmy rysunek pomocniczy: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/22642/hHKt1f5V%2B8byrIOK78Ofzg%3D%3D_6ad9ac619a46fcc80c479912d740729b8126f4c3c444a5f9e0efd030c8c800ec.png" width="305" height="206"> Wszystkie boki trójkąta ABC potraktujemy jako przeciwprostokątne pomocniczych trójkątów: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/22643/7o0HFz5YwtzhLcA%2B/w3dRw%3D%3D_4e3595686e77e40417ee95c29d2091a2e2c74ff70c78a2fc241476de1ce7c90e.png" width="305" height="206"> Policzmy teraz długości boków trójkąta ABC: ∣CY∣2+∣BY∣2=∣BC∣2 22+22=∣BC∣2 ∣BC∣=22+22<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.28em' viewBox='0 0 400000 1296' preserveAspectRatio='xMinYMin slice'><path d='M263,681c0.7,0,18,39.7,52,119
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