Resposta :
Considerando A o ângulo reto:
------------------------------------------------
B:
senB=AC/BC
cosB=AB/BC
tgB=AC/AB
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C:
senC=AB/BC
cosC=AC/BC
tgC=AB/AC
------------------------------------------------
B:
senB=AC/BC
cosB=AB/BC
tgB=AC/AB
----------------------------------------------
C:
senC=AB/BC
cosC=AC/BC
tgC=AB/AC
Explicação passo-a-passo:
=> Ângulo B
• [tex]\sf sen~\hat{B}=\dfrac{cateto~oposto}{hipotenusa}[/tex]
[tex]\sf \red{sen~\hat{B}=\dfrac{\overline{AC}}{\overline{BC}}}[/tex]
• [tex]\sf cos~\hat{B}=\dfrac{cateto~adjacente}{hipotenusa}[/tex]
[tex]\sf \red{cos~\hat{B}=\dfrac{\overline{AB}}{\overline{BC}}}[/tex]
• [tex]\sf tg~\hat{B}=\dfrac{cateto~oposto}{hipotenusa}[/tex]
[tex]\sf \red{tg~\hat{B}=\dfrac{\overline{AC}}{\overline{AB}}}[/tex]
=> Ângulo C
• [tex]\sf sen~\hat{C}=\dfrac{cateto~oposto}{hipotenusa}[/tex]
[tex]\sf \red{sen~\hat{C}=\dfrac{\overline{AB}}{\overline{BC}}}[/tex]
• [tex]\sf cos~\hat{C}=\dfrac{cateto~adjacente}{hipotenusa}[/tex]
[tex]\sf \red{cos~\hat{C}=\dfrac{\overline{AC}}{\overline{BC}}}[/tex]
• [tex]\sf tg~\hat{C}=\dfrac{cateto~oposto}{hipotenusa}[/tex]
[tex]\sf \red{tg~\hat{C}=\dfrac{\overline{AB}}{\overline{AC}}}[/tex]