Resolvendo:
I. Com base no triângulo maior, temos:
[tex]tg\:(\alpha +\beta )=\frac{11}{7}[/tex]
Para descobrir a tg β, calculamos:
[tex]tg\:(\alpha +\beta )=\frac{tg\:\alpha +tg\:\beta }{1-tg\:\alpha \:.\:tg\:\beta} \\\\\frac{tg\:\alpha +tg\:\beta }{1-tg\:\alpha \:.\:tg\:\beta} =\frac{11}{7} \\\\\frac{\frac{5}{7} +tg\:\beta }{1-\frac{5}{7} \:.\:tg\:\beta} =\frac{11}{7} \\\\\frac{\frac{5}{7} +tg\:\beta }{1-\frac{5\:.\:tg\:\beta }{7} } =\frac{11}{7} \\\\11\:.\:(1-\frac{5\:.\:tg\:\beta }{7})=7\:.\:(\frac{5}{7} +tg\:\beta )\\\\11-\frac{55\:.\:tg\:\beta }{7} =5+7\:.\:tg\:\beta \\\\77-55tg\:\beta =35+49tg\:\beta \\\\[/tex]
[tex]49tg\:\beta +55tg\:\beta =77-35\\\\104tg\:\beta =42\\\\tg\:\beta =\frac{42}{104} :\frac{2}{2} =\frac{21}{52} \\\\Portanto:\\\\tg\:\beta =\frac{21}{52}[/tex]
Assim, letra A.
Espero ter ajudado!
Desculpe qualquer erro.
Dúvidas, estou à disposição.
