Danielmihaelescask
a fost răspuns

ΔABC-dreptungic in A; sinC= 3 supra 5; AB+BC=24 aflati AB;BC;CA.

Răspuns :

sinC=AB/BC=3/5 ⇒ (AB+BC)/BC=(3+5)/5 ⇒ 24/BC=8/5 ⇒ BC=15 ⇒ AB=9;
AC²=BC²-AB²=225-81=144 ⇒ AC=12
Mariusdarius07ask


sinC=AB/BC=3/5

AB=24-BC

24-BC/BC=3/5

120-5 BC=3BC

120=8BC

BC=15

AB=9

AC=12

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