1)
b = 15/4 c = 8/3
Calculo de la hipotenusa a, por T. de Pitágoras:
a=√((8/3)^2+(15/4)^2 )=√(64/9+225/16 )=√(3049/144)
Calculo de las razones trigonométricas:
sen B=b/a= (15/4)/√(3049/144)=(15/4)/√(3049/144) √(3049/144)/√(3049/144)=(15/4)/(3049/144) √(3049/144)=540/3049 √(3049/144)
cos B=c/a= (8/3)/√(3049/144)=(8/3)/√(3049/144) √(3049/144)/√(3049/144)=(8/3)/(3049/144) √(3049/144)=384/3049 √(3049/144)
tg B=b/c= (15/4)/(8/3)=45/32
cosec B=1/(sen B)= 1/(540/3049 √(3049/144))=3049/(540√(3049/144)) √(3049/144)/√(3049/144)=36/135 √(3049/144)
sec B=1/(cos B)= 1/(384/3049 √(3049/144))=3049/(384√(3049/144)) √(3049/144)/√(3049/144)=3/8 √(3049/144)
cotg B=c/b= (8/3)/(15/4)=32/45
sen C=cosB= 384/3049 √(3049/144)
cos C=sen B= 540/3049 √(3049/144)
tg C=cotg B= 32/45
cosec C=secB= 3/8 √(3049/144)
sec C=cosec B= 36/135 √(3049/144)
cotg C=tg B= 45/32