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Răspunsuri la întrebare
Laturile se vor afla DOAR prin Teorema lui Pitagora. La punctul a) voi scrie mai complex Teorema si functiile trigonometrice, iar la celelalte voi scrie mai direct.
a) ΔABC-dr
sinB=cateta opusa\ipotenuza
=AC\BC
=16\20 (simplificam prin 4)
=4\5
sinC=cateta opusa\ipotenuza
=AB\BC
=12\20 (simplificam prin 4)
=3\5
cosB=cateta alaturata\ipotenuza
=AB\BC
=12\20 (simplificam prin 4)
=3\5
cosC=cateta alaturata\ipotenuza
=AC\BC
=16\20 (simplificam prin 4)
=4\5
tgB=cateta opusa\cateta alaturata
=AC\AB
=16\12 (simplificam prin 4)
=4\3
tgC=cateta opusa\cateta alaturata
=AB\AC
=12\16 (simplificam prin 4)
=3\4
ctgB=cateta alaturata\cateta opusa
=AB\AC
=12\16 (simplificam prin 4)
=3\4
ctgC=cateta alaturata\cateta opusa
=AC\AB
=16\12 (simplificam prin 4)
=4\3
b) ΔABC-dr
sinB=AC\BC
=24\30 (simplificare prin 6)
=4\5
sinC=AB\BC
=18\30 (prin 6)
=3\5
cosB=AB\BC
=18\30 (prin 6)
=3\5
cosC=AC\BC
=24\30 (simplificare prin 6)
=4\5
tgB=AC\AB
=24\18 (prin 6)
=4\3
tgC=AB\AC
=18\24 (prin 6)
=3\4
ctgB =AB\AC
=18\24 (prin 6)
=3\4
ctgC=AC\AB
=24\18 (prin 6)
=4\3
c).........
sinB=AC\BC=21\35
sinC=AB\BC=28\35
cosB=AB\BC=28\35
cosC=AC\BC=21\35
tgB=AC\AB=21\28 (prin 7)=3\4
tgC=AB\AC=28\21 (prin 7)=4\3
ctgB=AB\AC=28\21 (prin 7)=4\3
ctgC=AC\AB=21\28 (prin 7)=3\4
d)...
sinB=AC\BC=32\40 (prin 8)=4\5
sinC=AB\BC=24\40 (prin 8)=3\5
cosB=AB\BC=24\40 (prin 8)=3\5
cosC=AC\BC=32\40 (prin 8)=4\5
tgB=AC\AB=32\24 (prin 8)=4\3
tgC=AB\AC=24\32 (prin 8)=3\4
ctgB=AB\AC=24\32 (prin 8)=3\4
ctgC=AC\AB=32\24 (prin 8)=4\3
e)
sinB=AC\BC=27\45 (prin 9)=3\5
sinC=AB\BC=36\45 (prin 9)=4\5
cosB=AB\BC=36\45 (prin 9)=4\5
cosC=AC\BC=27\45 (prin 9)=3\5
tgB=AC\AB=27\36 (prin 9)=3\4
tgC=AB\AC=36\27 (prin 9)=4\3
ctgB=AB\AC=36\27 (prin 9)=4\3
ctgC=AC\AB=27\36 (prin 9)=3\4
f) ..............
sinB=AC\BC=40\50 (prin 10)= 4\5
sinC=AB\BC=30\50 (prin 10)=3\5
cosB=AB\BC=30\50 (prin 10)=3\5
cosC=AC\BC=40\50 (prin 10)= 4\5
tgB=AC\AB=40\30 (prin 10)=4\3
tgC=AB\AC=30\40 (prin 10)=3\4
ctgB=AB\AC=30\40 (prin 10)=3\4
ctgC=AC\AB=40\30 (prin 10)=4\3