Zadanie o trojkacie.
AB=12m
AC=10m
BC=15m
Ile stopni ma kat BAC?
Wydaje mi sie, ze trzeba uzyc wzoru na cosinus...
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15² = 10² + 12² - 2 * 10 * 12 * cosa
225 = 100 + 144 - 240cosa
240cosa = 244 - 225
240cosa = 19
cosa = 19/240
cosa = 0,0792
a ≈ 85°