Răspuns :
Construim inaltimea AD perpendicular pe BC
in ΔADB (<D=90°) <ABD=30°: cateta opusa < de 30°=ipotenuza/2
AD=AB/2=12√3/2=6√3 (inaltimea corespunzatoare ipotenuzei)
cos30°=√3/2 cos30=AB/BC (ΔABC)
AB/BC=√3/2
12√3/BC=√3/2 => BC=24
b) ΔADB: Teorema lui Pitagora : BD²=AB²-AD²
BD²=3X144-3X 36
BD=√324
BD=18 CD=24-18=6 ( proiectiile )
in ΔADB (<D=90°) <ABD=30°: cateta opusa < de 30°=ipotenuza/2
AD=AB/2=12√3/2=6√3 (inaltimea corespunzatoare ipotenuzei)
cos30°=√3/2 cos30=AB/BC (ΔABC)
AB/BC=√3/2
12√3/BC=√3/2 => BC=24
b) ΔADB: Teorema lui Pitagora : BD²=AB²-AD²
BD²=3X144-3X 36
BD=√324
BD=18 CD=24-18=6 ( proiectiile )
Construim inaltimea AD perpendicular pe BC
in ΔADB (<D=90°) <ABD=30°: cateta opusa < de 30°=ipotenuza/2
AD=AB/2=12√3/2=6√3 (inaltimea corespunzatoare ipotenuzei)
cos30°=√3/2 cos30=AB/BC (ΔABC)
AB/BC=√3/2
12√3/BC=√3/2 => BC=24
b) ΔADB: Teorema lui Pitagora : BD²=AB²-AD²
BD²=3X144-3X 36
BD=√324
BD=18 CD=24-18=6 ( proiectiile )
in ΔADB (<D=90°) <ABD=30°: cateta opusa < de 30°=ipotenuza/2
AD=AB/2=12√3/2=6√3 (inaltimea corespunzatoare ipotenuzei)
cos30°=√3/2 cos30=AB/BC (ΔABC)
AB/BC=√3/2
12√3/BC=√3/2 => BC=24
b) ΔADB: Teorema lui Pitagora : BD²=AB²-AD²
BD²=3X144-3X 36
BD=√324
BD=18 CD=24-18=6 ( proiectiile )