Problema 15 va rog,dau coroana

Răspuns:

tg=cateta opusa/cateta alaturata

tg(c)=[tex]\frac{AB}{AC}[/tex]

[tex]\frac{AB}{AC} =\frac{4}{3} \\[/tex]

rezulta AB=[tex]\frac{4AC}{3}[/tex]

Aplicam teorema lui pitagora

[tex]BC^{2} =AB^{2} +AC^{2}[/tex]

[tex]BC^{2} =\frac{16AC^{2} }{9} +AC^{2}[/tex]

aducem la acelasi numitor

9*225=25[tex]AC^{2}[/tex]

rezulta AC=9, AB = 12

A=[tex]\frac{C1*C2}{2} =\frac{AB*AC}{2} =\frac{9*12}{2} =54cm^{2}[/tex]

P=AB+AC+BC=12+9+15=36cm

Răspuns:

Explicație pas cu pas:

tg C=b/c=4/3

b²+c²=a²=225              16c³/9+c²=225     16c²+9c²=225×9

25c²=225×9      c=√225×9/25=15×3/5=9

b=9×4/3=12

P=a+b+c=15+12+9=36 cm

A=bc/2=12×9/2=54 cm²