Si el punto Q(8; 15) pertenece al lado final de un ángulo canónico β. Calcula: R = cscβ – cotβ
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Respuesta:
senA = y/r = 4/5
cosA = x/r = -3/5
tgA = y/x = 4/-3 = -4/3
ctgA = x/y = -3/4
secA = r/x = 5/-3 = -5/3
cscA = r/y = 5/4
Explicación paso a paso:
Sea "A" el ángulo.
P(-3; 3)
x = -3
y = 4
r = ?
Finalmente:
senA = y/r = 4/5
cosA = x/r = -3/5
tgA = y/x = 4/-3 = -4/3
ctgA = x/y = -3/4
secA = r/x = 5/-3 = -5/3
cscA = r/y = 5/4
Explicación paso a paso: