W kwadrat ABCD o boku 10cm wpisano trojkat rownoramienny DEF (DE=DF) W taki sposob ze E nalezy do boku AB a F nalezy do boku BC. Wiedzac ze pole trojkata wynosi 18 oblicz sinus kata EDF
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Pefd=|DE|²/2*sinα
sinα=36/|DE|²
P kwadratu −( 2*PΔAED +PΔEBF)= PΔEFD
100-(10*(10-x)+½x²)=18, x∈(0,10)
|DE|²- liczysz z tw. pitagorasa
sinEDF=9/41
liczę na naj