III gimnazjum

Policzmy to razem 3

Rozwiązanie 1

Do obliczenia pola trójkąta brakuje nam  długości podstawy AB. Oznaczmy sobie pomocnicze odcinki x i y.  <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31553/4D657eYdKEGdXyIVWUknmQ%3D%3D_11c8c7cb5b54be815315893e79e6b6981cc1a9d96de54a0c9db515981cca6567.jpg" width="501" height="297">   tg44o=x10​ tg44o=0,9657 x10​=0,9657             /⋅x

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31554/bQ1R%2BTTm8ygazK1YmInUOA%3D%3D_e397f6ac5a9f541bd21294edea80f0813e1173883e4258f8897502df12c6888d.jpg" width="750" height="370">   {x+40y​=tgαxy​=tgβ​   tgα=tg25o=0,4663 tgβ=tg55o=1,4281

Wyznaczamy wzór na długość przekątnej sześcianu. <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31555/0Htkt/6OHVCeiVmt3NEVPQ%3D%3D_c43eef52dc5a8f22b0902e16fc58a3a4532854241ab3a87c65dccadaa4ae346e.png"> Z twierdzenia Pitagorasa: a2+(2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
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